# Full text of "Mathematical questions with their solutions, from the "Educational times""

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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at http : //books . google . com/| Digitized by Google wnpi 6000302860 oogle Digitized by Google Digitized by Google MATHEMATICAL QUESTIONS, WITH THEIE SOLUTIONS. FEOM THE "EDUCATIONAL TIMES.' WITH MANT IPsptrs wxH Sahitaxts xmt irublis^eb in t^t '* (EbucHtbnal Sinus/' EDITED BT , W. J. MILLEE, B.A., MATBRMATIGAL MA8TBB, HUDDXB8FIBLD COLLKOR. VOL. XI. FROM JANUARY TO JUNE, 1869. LONDON: 0. p. HODGSON & SON, GOUGH SQUARE, FLEET STOEBT. 1869. Digitized byVjOOQlC *.* 5it* Volumes L to X. may still he hady price 6s, 6d. each. Digitized byVjOOQlC LIST OF CONTRIBUTOES. rris, Berks. s in the UniTenity of Cam- of France. r Office, Southampton. ibridge. d; raddent of the Queens- Gf^Ot Cambridge, riore di Milana ral Military Aoad., Woolwich. kjhool. Queen's University, Belfest. n the B. M. Acad., Woolwich. St. John's College, Sydney* . (Cadets') CoU., Sandhurst. iversity College, London. 50. ge, Cambridge, r Warrington. Grammar School. Digitized byVjOOQlC IV Lilosophy in the id Architectiire, >ridge. Dublin. ; Mathematical BTAJTLEY, ABCHES. Symbs, B. W., BJL, London. Stlyesteb, J. J^ F.B.S^ Professor of Mathematics in the Boral Military Acaden^, Woolwich ; Corresponding Member of the Institute of France. Tait, p. G.^V.A., Professor of Natural Philosophy in the Uniyersity of Edinburgh. Tasleton, FBAKCis A., M.A., Pellow of Trinity College, Dublin. Tatlob, C, M JL, Pellow of St. John's College, Cambridge. Taxlob, H. M.. B.A., Pellow of Trinity College, Cambridge; Vice-Principal of the Boyal School of Naval Architecture, South Kensington. Taylob, J. H., BA., Cambridffe. Tbbay, Septimus, BJl., Heaa Master of Rivington Grammar SchooL Thomson, P. D., M JL, St. John's College, Cambridge. ToDHUNTEB, IsAAC, P.B.S., St. John's College, Cambridge. ToMLiNSON, H., Christ Church College, Oxford. ToBEij.1, Gabbiel, Naples. TOBBY, Bev. A. P., MJl.. St. John's College, Cambridge. TowNSEVD, Bev. B., MA., P.B.S., Pellow of Trinity Colleffe, Dublin. TncKEB, B., M.A., Mathematical Master in University College School, London. WaIiEEB, J. J., MJL, Mathematical Master in Univeraity College School, London. Walmsley, John, Woolwich Common. Wabbbn, B., ma.. Trinity College, Dublin. Watson, Stephen, Haydonbridge, Northumberland. Whitwobth, Bev. W. A., M.A., Professor of Mathematics and Natural Philosophy, Queen's College, LiverpooL Wilkinson. Bev. M. M. U., Beepham Bectory, Norwich. Wilkinson, T. T., P.B.A.S., Burnley. Wilson, J. M., MA., P.G.S., Pellow of St. John's College, Cambridge; Mathe- matical Master in Bugby School. Wilson, Bev. B., D.D., Chelsea. Wolstbnholme, Bev. J., MA., Pellow of Christ's College, Cambridge. WooLHOUBE, W. S. B., P.BA.8., &c., London. CbfUributors deceased since the PubUeation qf Vol. I, Db Mobgan, G. C. ma.: Holditch, Bev. H., MA.; Lea, W.; O'Callaohan, J. ; PuBKiB8,^H. J., BA. ; Pbouhbt, E. ; Sadlbb, G. T., P.BA.S.; Wbioht, Bev. B. H., M.A. Digitized byVjOOQlC CONTENTS. iWatfttmatiral ^aptrsf, ^r* • No. Page 70. To find the number of permntations of n things taken r to- gether. By C. R. RiPPiN, M.A 40 71. Note.on Question 2740. By Professor Oailby 49 72. To express the distance between the centres of the circmnsoribed and inscribed circles in terms of the radii of those circles. By J. Walmslet, B.A 63 73. Note on the late Judge Hargreave's Solution of the Quintio. By the Rev. T. P. Xi&kkan, M.A., F.R.S 82 1587. Apollo and the Muses accepted the challenge of Jove, to vary the arrangement of themselves on their filzed and burnished couches at his evening banquets, till eveiy three of them should have occupied, once and once only, every three of the couches, in every and any order. In how many days, azid how many ways, did they accomplish the feat, keeping one arrangement of themselves through all the solutions P Re- quired two or more of these solutions, clearly indicated, so as to save space, by cyclical operations 65 1819. From two fixed points on a given conic pairs of tangents are drawn to a variable confocal conic, and with the fixed points as foci a conic is described passing through any one of the four points of intersection. Show that its tangent or normal at that point passes through a fixed point 83 1878. If a line of given length be marked at random in n points, and broken up at those points, find the chance that the sum of the squares on the parts shall not exoeed one-nth of the square on the whole line 17 1912. If a and b be the points of contact, with a curve of the third class, of a double tangent ; and if this tangent be intersected in m, n, p by the three tangents to the curve which can be drawn from any point in the plane, then , — '— — 1^ = — > hn,bn ,hp p^ where p^, p^ are the radii of curvature at the points a and 6 41 Digitized byVjOOQlC VI CONTENTS. No. Page 2101. Trouver leB conditions n^oeesaireB et soffiBantes pour que lea qoatre raoinea d'one ^nation dn qnotri^me degr6 forment un qnadrilat^re inscriptible. Trouver la aarface et le rayon de oe qnadrilai^re 46 2108. Required analogaea in Solid Gteometry to the following pro- positions in Plane Geometry : — (a,") The perpendionlara of a triangle meet in a point. (b.) The middle points of the diagonals of a quadrilateral are in one straight line. (c.) The circles whose diameters are the diagonals of a quadrilateral have a common radical axis. (d4 Every rectangular hyperbola circumscribing a triangle passes through the intersection of perpendiculars. (e.) Every rectangular hyperbola to which a triangle is self-conjugate passes through the centres of the four touching circles. (f.^ sin(A + B) BsinAcosB + cosAsinB. (gr.) The sum of the angles of a triangle «s two right angles. (h.) In any triangle !mA ^ rinB ^ mnO ^^ a c 2109. Two lines are drawn at random across a convex dosed curve ; determine the chance of their intersecting 94 2221. 1. Show how to determine the locus of the feet of perpen- diculars from a fixed point on the generating lines of a system of confooal quadrics. 2. Prove that this locus is described by foci of the plane sections passing through the fixed point 64 2244. Form 11 symbols into sets, 6 symbols in a set, so that every combination of 4 symbols shall appear once in the sets 97 2801. A circle is drawn so that its radical axis with respect to the focus S of a parabola is a tangent to the parabola ; if a tan- gent to the circle cut the parabola in A, B, and if SG, bisect- ing the angle ASB, cut A3 in C, the locus of C is a straight line 31 2345. Determine in trilinear coordinates the foot of the perpen- dicular from the point (a/, y', Z) on the line Ix + my + nz ^^0; also the reflexion of the same point with regard to the same line 30 2369. Given a cubic curve K, and a point on it p; through P is drawn any transversal meeting E again in m and n, and on it is taken a point x such that the anharmonic ratio (jpTnvn) shall be equal to a given quantity. Prove that the locus of is a quartic curve with a point of osculation at p, touching K at j7 (counting for four points of intersection) and in four other points 43 2456. If a, b, c be the three sides of a spherical triangle, and k the radius of its polar circle, prove the formula ^^j , _^ (cos h cos c sec a— 1) (cos c cog a sec b— 1) (cos a cos h sec c— 1) ^^ 4 sin « sin {s—a) sin (s— b) sin («— c) Digitized byVjOOQlC CONTENTS. Vll No: Page 2456. Show that i£ an ellipse pass through the centre of a hyper- bola, and have its foci on the hyperbola's asymptotes j (a) the hyperbola passes through the centre of the ellipse, (b) the axes of each curve are respectively tangent and normal to «ach other, and (c) the two axes which are also normals are equal to each other 31 2486. Pind a point on an ellipse such that the normal produced to meet tbe curve again may cut off a maximum area ^. 91 2494. On each side of a hexagon as base, a triangle is described by producing the two adjacent sides to meet, and a second hexa- gon is formed by joining the vertices of these triangles in order. Show that if either of these two hexagons can be in- scribed or circumscribed to a conic, the other can be circum- scribed or inscribed to a conic 44 2513. Let A, B, 0, D, B, F be six points. Let BF, CD meet in a ; AB, CB in b; AB, CD in c; AF, CB in d; BF, DE in e; and AF, DB in /. Show that, if either of the hexagons ABCDBF, dbcd^ can be inscribed or circumscribed to a conic, the other can be circumscribed or inscribed to a conic 44 2525. Three equidistant lines are drawn parallel to an asymptote of a hyperbola, and a triangle inscribed where the lines meet the curve. Then any line parallel to the other asymptote will be divided harmonic&lly by the three sides of the triangle and the curve t 20 .2527. Find the locus of the intersection of two tangents to a circle, the chord of contact of which subtends a right angle at a fixed point ; and also of the middle points of a system of chordB of a circle, which subtend a right angle at a fixed point; and thence show that the envelope of a system of chords of a conic, which subtend a right angle at the focus, is another conic of coincident focus ; and that the locus of the middle points of a system of chords of a conic which sub- tend a right angle at a fixed point is another conic 93 2533. Find the envelope of a line upon which two given conies inter- cept segments which have a common middle point ; and find also the locus of this middle point 84 2539. If three conies cut one another so that every two of them have a common chord which is also a chord of a fourth conic, the other three chords of intersection of the three conies meet inapoint .■ 44 2544. There are two values of d for which o, jS, 7, the roots of the cubic agfi + Sha^ -k- 3cx + d a fulfil the linear relation Aa-l- B/3+ C7 B 0. Show how to find the quadratic equation of which these values of d are the roots ^ 64 2597. A right cone, whose weight may be neglected, is suspended from a point in its rim ; it contains as much fiuid as it can : show that the whole pressure upon its surface is , ,. sinacosd (cos(0-i-a)f$ Digitized byVjOOQlC viii CONTENTS. No. Ft«e where h and 2a are the height and vertical angle of the cone, and 9 is determined from 3 sin 20 » 4 sin 2(0— a) 101 259S. A nniform rod (mass «* fi) is placed inside a spherical shell (ma8s» m, radios ^ a), which, under the influence of gravity, rolls down the surface of a rough sphere (radius » r) ; find equations for the movements of the rod and shell, there being no firic^ion between them ; and hence show that, if a particle /K, is placed near the lowest point of a spherical shell (m) which performs small oscillations on a rough horizontal plane, the length of the simple equivalent pendulum is a 62 2m + Ai In a plane, (1) on every line, there are two points harmonic conjugates to the conies through four points A, 6, G, D. They are real, if the product of the four triangles ABC, BCD, CDA, DAB, and the perpendiculars firom the points A, B, C, D on the line, is positive. Tf a line does not cut a conic, why are the conjugates on the line real, to any set of four points on the conic ? Give a rule for determining the nature of the points on a line, when the product of the triangles is (1) positive, (2) negative. (2.) Through any point, there are two lines conjugate to the conies inscribed in four lines. They are real in those regions of the plane formed by the four lines in which the product of perpendiculars from the point has the same sign as that of points in the conVex quadrilateral. Hence every point on or within a conic inscribed in the four lines lies in these regions 106 2606. The developable circumscribing two surfaces of the second degree touches either of them along a curve, which is its intersection with another surface of the second degree 68 2616w Let two intersecting tetrahedra have all their edges bisected by the same system of Ciurtesian axes, each axis through two opposite edges of each tetrahedron ; then the solid about the origin has the origin for its centre of figure 47 2618. A pack of n different cards is laid, face downwards, on a table. A person names a certain card, that and all the cards above it are shown to him and removed ; he names another, and the process is repeated. Prove that the chance of his naming the top card during the operation is i_l + l_J.+ -.(nil" 70 2622. The equation connecting the distances (ri, r2, rj) of any point on a Cartesian oval from the foci is (/3-7)o*ri+ (7-a) /bVj+ (a-0) 7^3 = 0, where a, /3, 7 are the distances of the foci from the triple focus 66 2623. If a conic pass through the four points of contact of tangents to a cubic from a point (A) on the curve, and through two other points (B, C) on the cubic ; then A is the pole of BC with regard to the conic 68 Digitized byVjOOQlC CONTENTS. IX No. Page 2636. 1. If A, B be two confooal oonics, and A! be a third oonio having doable contact with A, show that there exists a fonrth conic B' having double contact with B and confocal with A'. 2. If A, B be two confocal quadrics, and A' be a third qnadric having continnous contact with A, show that there exists a fourth qnadric B' having continnons contact with B and confocal wi^ A! ^ 43 2640. I. If a conic of given eccentricity e cironmBcribe a triangle ABC, the locns of its centre is {yz sin» A + gg sin^ B + a?y sin^ C)^ ^ (fr^. xyz (w + y + z) sin^ A sin* B sin* C 1— e* ' and (2) if it be inscribed in the triangle, the locns of the centre is (jF* sin* A+ ...+2pz sin B sin C cos A-f ...)* _ 4(e*- 2)* . xyz (x + y + z) sin* A sin* B sin* "" 1— e* the triangle of reference having its angular points at the middle points of the sides of the triangle ABC. 3. Hence show that the locus of points with which as centre two similar conies can be drawn, one inscribed and the other circumscribed to the triangle, is two circles, one the circum- scribed circle of the triangle, and the other that which haa the centre of gravity and the centre of perpendiculars of the triangle as the ends of a diameter 25 2642. If three surfaces of the second degree meet in four points, at each of which their three curves of intersection have a com- mon tangent, these four points lie in one plane 77 2651. Determine the form of the solution of the differential equation ^•^^'^S * '•'"^'^^ "" {•^'<'>±"'} ^ = <* ^^ 2653. 1. If the centre of one of the four circles which touch the sides of a triangle self-conjugate with respect to a parabola lies on the directrix; show that its circumference passes through the focus. 2. & the point of intersection of the three perpendiculars of a triangle inscribed in a parabola lies on the directrix $ show that the self -conjugate circle of this triangle paases through the focus of the curve 32 2654. Determine the polar reciprocal of the quartic curve y«a;* + »*«*-H«^* + 2ay«(l» + wiy-»-w») «0 60 2670. An observer, seated in the aisle of a cathedral and looking westward, sees the horizontal lines above the arches con- verging to a point before him, and consequently cutting the horizon at a certain angle. On his facing northwards, he sees that the same lines are now parallel to the horizon. Bequired the curve that they will appear to lie in, as he gradually turns his head through 90° 57 2688. A variable triangle circumscribes an equilateral hyperbola, and is such that its nvne-^pomt circle passes through the centre of this curve ; prove that the locus of the centre of its cirCMwwonlmi^ circle is the hyperbola in question «., 59 b 7-' Digitized byVjOOQlC :X CONTENTS. No. Fkffe 2701. In a right conoid whose axis is the axis of «, show that the radii of principal cnrvatnre at the point (r cos 9, r sin 9, «) are giren hj the equation v(S)V-»S{"*(S)T-!'"-(S)T '" 2712. Form 15 symbols into sets, 5 symbols in each set, so that every combination of 4 symbols shall appear once in the sets 97 2718. Find in plomo the loons of a point F, snoh that firom it two given points A, C, and two given points B, D, subtend eqnal angles 83 2719. 1. The six points which, in conjunction with any common transversal, divide harmonically the six sides of any tetra- stigm, lie all on one conic, which also paases through the three points of intersection of opposite sides of the tetrastigm. 2. Of the six points, if those pairs which lie on opposite sides of the tetrastigm be joined by straight Unes, the three straightlines thus drawn are concurrent, and their point of inter- section is the pole of the transversal with respect to the conic 21 2724. Three smooth rings, F, Q, B, are thread on an endless string of given leng^, and constrained to move on three straight rods, OF, OQ, OB. To investigate the motion oorrespondmg to a slight arbitrary disturbance of the system 75 2726^ If a conic touch the sides of a triangle and pass through the centre of the circumscribed circle, the director circle of the conic will touch the circumscribed circle of the triangle 60 2727. Two lines (a) and (h) in space touch a fixed quadric U, show that a variable line intersecting the two former and touching the latter generates another quadric 20 2728. Given three planes and their poles with regard to a system of quadrics, the locus of centre js a right line 101 2736. Find the curve whose radius of curvature and radius vector at each point are equal to one another, and prove that it is the second involute to a circle having an apse midway be- tween the cusp of the first involute (from which it is derived) and the centime of the circle 18 2740. Prove that the envelope of a line given by an equation of the form L cos 2^ -i- M sin 2^ + P cos ^-t-Qsin^-t-B » may be obtained as the discriminant of a cubic equation, and form this equation 42 2741. A moveable event (depending on the moon) happened in the year y, show that it wiU happen again in the year y + 19 (28t— 5a + 86) ; where amay be 0, 1, 2, 3 j h the num- ber of completed centuries since the year y which are not leap years; and t any arbitrary integer 52 2745. Show that the average area of all the ellipses that can be in- scribed symmetrically in a portion of a parabola cut off by a double ordinate perpendicular to the axis, in parts of the area of the parabola, is -^w if (1) the centres be equably distri- buted over all possible positions on the axis, and ^ir if (2) Digitized byVjOOQlC CONTENTS. XI No. Fige the intersections of the ohords of contact with the aoda be equably distribnted on the axis 24 2756. Form 16 symbolB into sets, 4 symbols in each set, so that eveiy triad in the symbols shall appear once in the sets 97 2757. If ^^j.^.^i ». 9, 1 (1). (2). show that each of the equations a« (ar-a?o)» + b« (y~yo)» ^ a^ (^-0^0'+ b» (y-yi)» (a?a?o+yyo-l)' " {xx^-i- ^f^-iy o^(x~aro)8 + ba(y~yo)8 ^ aH^-arja + b' (y^yQ' (a3fo-a'cy)«-(af-aro)'-(y-yo)* * (ayi-ariy)*- (af-a?i)»-(y-yi)«*" represents the right line L*-0 and a cubic curve 83 2761. Find the conic envelope of the radical axis of two circles, one of which is fixed, whilst the other passes through a fixed point. If the radii of the circles are equal, find the foci and axes of the envelope 74 2762. Find the envelope of an ellipse which has one vertex in the curve of a given parabola, and touches at its adjacent vertex the extremity of a fixed double-ordinate of the parabola 58 2763. Give a geometrical interpretation for the relation between the invariants of two conies, viz., ee' » AA' 94 2764. It is weU known that if A, P, P' be fixed points on a line (that is, if the distances AP, PP' are constant), and if PP' describe given fixed lines, say the directrices, then the point A de- scribes an ellipse having for its centre the intersection of the two directrices ; and the question hence arises : Given three positions of the point A and of the generating line APP*, to find the directrices ; or, what is the same thing. Given the lines 1, 2, 3, passing through the points A, B, G respectively, it is required to draw & line cutting the three lines in the points P, Q, B respectively, such that AP » BQ = OK 19 2766. Three points move from a fixed point in three different but given directions, with given uniform but different velocities. Find the locus of the centre of the circle passing through them 22 2769. From a point, taken at random within a triangle, perpen- diculars are drawn on the sides ; find the probability that the triangle formed by joining the feet of these perpendiculars is acute-angled 94 2770. If a conic be drawn through S, S', the foci (both real or both impossible) of a given conic, and osculate the conic at a point P, the tangents at S, S' will intersect in the centre of curva- ture at P : 60 2771. Prove that (1) if a straight line of length a be divided at random in two points, the mean value of the sum of the squares on the three parts is \o? ; (2) if the line be divided at random into two parts, and the longer part again divided at random into two parts, the mean value of the sum of the Digitized byVjOOQlC Xll CONTENTS. No. Ffege squares on the three parte is ffo^ ; and (3) if it is an even chance that n times the sum of the squares on the parts in (1) is less than the square on the whole line, then »-i7TT73 ^« 2773. When a given angle rolls upon a fixed parabola, the locus of its vertex is well known to be a hyperbola having the same focus and directrix as the parabola. Conversely, when a given parabola rolls within a fixed angle, show that its focus and vertex describe lines of the fourth and sixth orders re- spectively; find the actual equations of both, these curves, and thence show that the vertex of the fixed angle is a conjugate point on the former curve, and that the equation of the latter, when the given angle is a right angle, becomes a^ (a^ + y' + 3a?) — a« or x^y^ (a?* + y*) « a?, the sides of the given angle being the axes, and a equal to the distance of the focus from the vertex 9$ 2774. A central conic {h^a^+ahf^ «= aPh^) is turned in its own plane about its centre through a right angle ; prove that the locus of the intersection of the normal at any point on the given conic with the same line in its new position is one of the two sextic curves (a^ + t/^^{a''(ai±yy±h\xTyy] « (a«+b3)'(«»-y')* 86 2776. Through A, the double point of a circular cubic, draw AB perpendicular to the asymptote ; if chords be drawn to the curve subtending a right angle at the double point, show that there is a fixed point in AB at which also they subtend a right angle 63 2778. Suppose fjL any quantity fractional or integral, and write C-«,;C-^ c^^ttSapa^ ^. MO^^^iy^X .- also let 71 be a positive integer such that /i and /jL + n hitve not different signs ; then it has been proved elsewhere, and may be assumed to be true, that the number of changes of sign in the series a^, fltl' — OoOj, Og^— OiOj, is em inferior Umdt to the number of imaginary roots in the equation c^* + cia?""* + <j2a?*"*+ — 0. 1. Bequired to prove that the above theorem remains true when we write <,-a,. c^-^o,, c,-f^P^<H. c,-dt±Mt±^a, , y y{y-i-l) y{v+l)(y + 2) provided y is a positive integer, and ^— v+1, fjL + y do not differ in sign. 2. Ascertain whether y being an integer is essential to the truth of the extended theorem 38 2779.- Show that the famous equation between wrc, angle of cou' Digitized byVjOOQlC CONTENTS. Xin No. Page Ungence, and perpencUciua/r on tcmgent to a corye, viz., I> + j~ » -J-* may be interpreted aa affirming the self -evident proposition that the radius of curvature is the distance between the tangent at any point of a cuire and the tangent at the cor- responding point of its second evolute ; and account for the connecting sign between p and -^ in the above equation hemg phis Q3id not nvinus 21 2780. The envelope of the diord common to an ellipse and its circle of curvature is a curve of the fourth class which has three double tangents, one at infinity and the two others coincident with the conjugate diameters equally inclined to the axis. Prove this, examine the curve, and consider the cases of the hyperbola and parabola 62 2781. Prove that the maximum value of the common chord of an ellipse and its circle of curvature, a and h being the semiaxes of the ellipse, is j-^^±_^{2(a«-a»6»+M)i-(a» + l.»)(2o?-b»)(ab»-a?)}* 28 2782. 1. A given point is known to be within a certain circle of given radius, but unknown position; find the chance that another given point is also within the circle. 2. Three given points are known to be within a certain circle, which is otherwise altogether unknown; determine the most probable position of its centre. 3. Two given points are known to be within a circle, and a third given point is known to be outside it ; determine the most probable position of its centre 107 2787. The value ofZ/T x^dx dAj da taken over the whole volume of a tetrahedron through one of whose comers the plane of (yz) passes, is ^V (oj^+bi^ + Ci^), where V is the volume, and Oi, &i, Ci are the distances of the middle points of the opposite I from the plane of reference 110 2789. If two circles pass through the foci of an ellipse and touch the same variable tangent to the curve, the angle at which they intersect is constant and equal to 2 tan~^ - 62 c 2795. Find the average jK^uare of the distance between the centres of the inscribed and circumscribed circles of a triangle in- scribed in a given circle 64 2797. In any complete quadrilateral, the radical axis of the three circles whose diameters are the diagonals will pass through the centres of perpendiculars of the four triangles formed by the four straight lines 65 2798. To find the condition that a given point (x, y, z) may be ex- terior to the given conic S = (A, B, 0, F, G, H][ar, y, «)« - 41 Digitized byVjOOQlC XIV CONTENTS. No. Page 2799. If n dioe are tlurown, what is the chanoe of an odd number of aoes tnmuig up P 61 2800. Three oiroles paasing through a point P form a oironlar tri- angle ABC, and each side of this oironlar triangle or ifcs continuation is cut orthogonallj at the point P' by a cirole passing through P; prove that the three circles described about the triangles PAF, PBP", POP"' are coaxal 69 2804. A straight pole stands verticaJly on a slope inclined to the south. If it be broken at random by the wind blowing in a given direction, so that the upper end of the pole rests upon the slope, determine the probable area of the triangle thus formed j and deduce the result for a horizontal plane 69 2808. Four fixed tangents are drawn to a conic S; three other conies are drawn osculating S in any the same point P, and each passing through the ends of a diagonal of the circum- scribed quadrilatersJ : prove that the tangents to these conies at the ends of the diagonal meet in one point P', and that the locus of P' is a curve of the sixth degree and fourth class, having two cusps on every diagonal, and touching S at the points of contact of the four tangents 108 2813. The equation of a conic referred to an axis and tangent at vertex being aa^ + hy^ + 2dsD » 0, if the conic be turned about the vertex in its own plane through a right angle, the locus of the intersection of any tangent in the original position with the same line in the new position of the conic is the (bicircular) quartic l){a(«2 + ya)+2d(a?+y)} (x^ + y^ ^^{x^y)\ and the corresponding locus for the normal is the sextic What does this latter equation become in the case of the parabola? 71 2815. A cask contains a gallons of wine. Through a hole in the top water or wine can be let in at the rate of b gallons per mi- nute ; and through a pipe in the bottom, when open, the mix- ture can escape at the same rate. Suppose the discharge pipe is opened at the same instant that water is let in at the top, and t minutes afterwards the water is shut off and wine let in. Eequired the quantity of water in the cask at the end of ti minutes from the opening of the discharge pipe, and the length of time elapsed, both before and»afber wine was let in at the top, when the quantities of the two fluids in the cask were equal, supposing them to mingle perfectly 87 2816. Find three square numbers in arithmetical progression, such that the square root of each (a) increased or else (js) di- minished by unity shall give three rational squares 88 2818. Given a circle S and a straight line a not meeting S in real points ; O, O' are the two point-circles to which, and S, a is the radical axis ; two conies are drawn osculating S in the same point P, and having one focus at O, O' respectively : prove that the con esponduig directrices coincide 74 Digitized byVjOOQlC CONTENTS. XV No. Fftgo 2821. Show that the mean ralne of the distanoe from one of the foci of all points within a given prolate spheroid is ^(3 + 6^), 2a being the axis and 6 the ecoentrioity 87 2828. Show that on a chess-board the chance of a rook moving from one square to another without chang^g colour is f ; but that witliout altering the equality of the rmmher of the black and white squares, but only the manner of their distri- bution, the chance may be made equal to i 4B 2828. A straight rod is divided into n parts in arithmetical progpres- sion, and equal particles are fixed at the points of division. If the system be made to vibrate about one extremity, deter- mine the average length of the pendulum, neglecting the weight of the rod 83 2835. In a plane, if A, B are two points, and a point P describe a curve of the nth order, show that F', the intersection of the perpendiculars in the triangle PAB, will describe a curve of the 2nth order with three multiple points of the order n. Explain why the curve corresponding to a circle through the points A and B includes its reflexion to the line AB 86 2839. In a triangle the bisector of the base is equal to the less side and also to one-half of the greater side j determine the three angles c 73 2847. Prove that •»D-.,.*«-^ (l + 4,Wk)-j,-rTiXS^ where D - ± 67 dx 2870. Given three points A, B, G, and a conic j two points P, Q are taken on the conic such that the pencil A (BPQGJ is har- monic ; prove that the envelope of PQ is a conic touching AB, AO at points on the polar of A with respect to the given conic 83 2875. Any tangent to a conic is, of course, divided in involution by three other tangents, and the lines joining their points of intersection to one of the foci of the conic ; prove that the distance between the double points of the involution subtends a right angle at the focus of the conic. The locus of these double points, the three tangents being fixed, is a cubic having a double point at the focus, whose nature I have not examined 106 2877. If u^x + y + z + t, V *^fiD + gy + ha + kt, V wmygt + etw+txy + xyz, Y ^fyxt+gstx+htxy + kxyzi prove that the resultant of u, v, U, V is if-g) (f-h) (f-k) ig-h) (g-k) {h--h) xif+g-h-kyif+h-g-kyif+h^g-^h)^ 81 2887. 1. If a point P on a conic be connected with any two fixed points A and B in its plane, all chords which are divided harmonically by PA and PB will be concurrent. 2. The locus of the point of concurrence when P is variable Digitized byVjOOQlC XVI CONTENTS. No. Page is another oonic which has double contact with the g^yen one onAB 100 2893. If there be {n) quantities a, b, c, d, &'c., each of which takes independently a given nmnber of valaes aiagag. . . . b^bsbs. • . . &o, (the nmnber may be different for each), if we pat 2 (a) ^a+h + c + d + Ac, and if for shortness we denote "the mean value ofx" by M (0), prove that M p (a)] = M (a) + M (b) +M (c) +&c. « 5[M (a)], M[2(a)]3« {2[M(a)]}»-.2[M((i)]«+2[M(a>)] 99 2895. At any point on a cusped cubic a tangent is drawn meeting the cubic in a second point ; from the first point a Une is drawn touching the cubic in a third point. Prove that the lines drawn from the cusp to these points form with the cuspidal tangent a harmonic pencil 104 2898. A circle rolls with internal contact on a fixed circle of half its radius ; prove that the envelope of any chord of the rolling circle is a circle which reduces to a point for a diameter 105 2902. The circle passing through three points on a parabola, the normals at which co-intersect, always passes through the vertex; and if the point of co-intersection of the normals describe a coaxial conic having the vertex as centre and axes in the ratio of w : n, the locus of the centre of the circle will be a conic having the focus as centre, and axes in the ratio of 2w : n 104 Digitized by VjOOQIC MATHEMATICS VBOM THE BDtrOATIONAL TIMBS, WITH ADDITIONAL PAPBKB AND SOLUTIONS; 1878. (PMposed by W. E. CuBVOVD, B.A.)— -If a line of given length be marked at random in n points, and broken np at those points, find the chance that the sum of the squares on the parts shall not exceed one-nth of the square aa the whole line. SohOion hy the Bev. J. Wolbtevholmb, MA.. Let «!, r» . • . . «,^ be the distances, in order of magnitude, of the n points of division from one extremity, and 1 the length of the line; then the limiting equation is «f + (««-*i)* + • . . . (l-^n)" < i or a?ft.... +a'i-*iar,-.... -.«^.ia?^+ 't±} < 0. Now I assume as obvious that a point (in space of n dimenaons) whose coordinates (rectangular or not) are Xi,Xf, ..•, x^, is a random point when 4^1, jTj, .... dT^ are random lengths, and that the diance required is the value of ///'• • • • dxi dx2 .... dx^ when the limits are given by the equation a?! + ....+ar*-a?i»j-....+l!±l = (A), divided by the same integral when the limits are given by d?i < jr2 < d^ . . • • <x^<l, all positive. First, to find the quad-volume of the quadric (A). Its quasi-centre is given by Zxi^x^ « 2x^—Xi'-x^ » 2a?j— a?2~^4 =*•••• •= 2jf^— a?^_^— 1 » 0, 1 2 or flTs « 2*1, JTs B SoTi . . . . , or a?i « — -, a?j ^ .... YOL. XI. B Digitized byVjOOQlC 18 The equation of the surfiice referred to its eentre is then + «n-«l*2-«-«'-'ii-l« 2i»(ii-l-l) Now, transforming the axes (and to fiicilitate matters here I aasnme rectangular axes) so as to destroy the terms involving products, the resulting equation is «iXi + aaXa+ .... +"ji^« " ^JTJ^' where (see my Book of Mathematical PkoUems, Quest. 921) ai, o^, . . . . a„ are the roots of the equation 3:-.2, 1, 0, 0, 1, «-2, 1, 0» 0, .. 0, 1, «-2, 1. 0, , 0, 0, 0, 0, 0, 0, .1. «-2, 1 .0, 1. «-2 = 0, the discriminant of (arj+ .... +a?")«-2(a?* + .... +dr"-a?iai-.. ..), and the product of these roots is » + 1. Also the value of fff,.,^y,„ rfX„^ where the limits are riven hy oiX?+ .. .. +o_ X„ ■» — - — -. is (Tod- * " fi(ft + l) hunter's Integral Calculus, Art. 274) 1 f X )* * 1 ^ 1 _ S IT |*"__JL_ . (0,03 ...oj* «»(« + !)) r(irt+l) (»+l)M»(n + l)) r(i»+l)' and Jf^. . . . dxidx^ .... dir^, over all such positive values that arj < a?2 < «» • . . . < a?« < 1, is i or — i — ,. The chance required is then * [« r(n + l) 4 5 IT l*» r(» + l) (»+i)* «»(»+!)) r(i»+i)' [The Proposer's solution is given on p. 83 of Vol. VI. of the Seprint.J 2736. (Proposed hy Professor Stlvesteb.)— Find the curve whose radius of curvature and radius vector at each point are equal to one another, ahd prove that it is the second involute to a circle having an apse midway between the cusp of the first involute (from which it is derived) and the centre of the circle. Solution hy the Ret. James Wuite, M.A. The property of the second involute of a circle stated in this Question, viz., that when its apse is midway between the cusp of the first involute (from Digitized by Google 19 wbicb it 18 derived) and the centre of the circle^ its ndias vector and radius of carvatare will be equal, may be proved geometrically, as follows : — Let V be any p<nnt on this second involute whose apse is midway between O and B : it is required to prove that OV-PV. The second involute which commences at B, the cusp of the first, will cut PV in H, making HV»irr, (a being radius of the circle). Let QP=f • and draw OS perpendicular to VP; then OS » i, and SP * a. Now OV> = 0S« + VSa « ^+(VP-a)«, and raroBP + 4a =^ +?. VP = HP + ^ therefore OV = £1 + ? « VP ; 2a * 2 that is, radius vector (OV) » radius of curvature (VP). [Mr. Whitb states that the mam part of this proof is due to Mr. Cboftoh.] From this a simple method of finding a point on the second involute cor- responding to any one given on the first may be derived. Bisect the line joining the given point with the centre of the circle, and at the point of bisection erect a perpendicular. Also at the given point erect a perpendicular to the tangent from it to the circle ; the intersection of these perpendiculars is a point on the second involute which has its apse midway between the cusp of the first and the centre of the circle (the triangle OVP being isosceles). Take a point (H) on the perpendicular to the tangent distant from the one found one-half the length the radius of the circle, and it is a point on the second involute which commences at the cusp of the first. [Professor Sylyzsteb sends the following remarks on this question :— > The envelope of a set of half-pitch second involutes to any given circle is a drcle concentric with the first. Consequently anjf circle corresponds to a singular solution of the clifierential equation which defines the curve in question; and thus we obtain another and obviously true solution of the question.] 2764. (Problem by A. Smith, Q.C.)— It is well known that if A, P, P' be fixed points on a line (that is, if the dirtances AP, PF are constant), and if P, F describe given fixed lines, say the directrices, then the point A describes an ellipse having for its centre the intersection of the two directrices ; and the question hence arises : Given three positions of the |X>int A and of the generating line APP', to find the directrices; or, what is the same thing. Given the lines 1, 2, S, passing through the points A, B, C respectively, it is required to draw a line cutting the three lines in the points P, Q, B respec- tively, such that APsBQaCB. Digitized by Google 20 SoliOkM hff Abohbk SsAiruET. Let p, qfT be any three correspondiDg points on the three given Hnes ; that is to say, let the three segments Ap, Bg, Cr (set off in assigned direc- tions) be equal to one another. Then the connector pq will, by a well4Knowa theorem, envelc^ a parabola of which Ap is a tangent ; in like manner, the connector pr will envdope another parabola having the tangent Ap in common with the first. The remaining two common tangents of these para- bolas will manifestly be the two required directrices. The problem is thus reduced to the well-known one (soluble by rule and compass; of finding the common tangents to two parabolas which touch a given line. 2525. (Proposed by G. O. RAKLO]!r.)—>Three equidistant lines are drawn parallel to an asymptote of a hyperbola, and a triangle inscribed where the lines meet the curve. Then any line parallel to the other asymptote will be divided harmonically by the three sides of the triangle and the curve. Solution by the Rev. J. Wolstenholmb, M.A. ; the Psofobxb ; €md othen. First to find the locus of a point on any line in a given direction forming a harmonic range with the points where it meets the sides of the triangle of reference. If (/ + Ar) a? + (»i + ifc) y + (n + Ar) « = be any such straight line, then for the point (P) in question we have («i + Ar)y = (»+A?)a (if A'B'CP be harmonic), and the locus is . (Za? + «^ + n2)(y-«) + (» + y + 2:)(»«-«iy)=0 (1); or, in another form, 2(wi— »)y«— (»— Qza?— (?— «i)a?y = (2). Equation (2) shows that this is a conic circumscribing the triangle, and (1) that the asymptotes are parallel to lx-^m^-¥nz^Q and to y»«; the latter of which being the bisector of the side BC, shows that the three straight lines through A, B, parallel to one asymptote are equidistant. Hence the truth of the theorem. 2727. (Proposed by the Rev. R. Townsend, P.R.S.)— Two lines (a) atid {h) in space touch a fixed quadric U, show that a variable line intersecting the two former and touching the latter generates another quadric. Solution hy W. S. Bubnsidb, M JL. It will be convenient to write the equation of the quadric IT under the form aa^ + hy^ + 2/nxy + 2trzw » 0, where the planes z and w plainly touch the surface, and the line (a?, ^) is their chord of contect. Whence we may consider {x, z) and (y, w) as the tangent lines (a) and (b). This being so, the condition that u variable line x =^\z, » » juy, should touch the quadric IT is ab\^ = (n\ + fwy, giving for the surfiuse traced out by it the equation abo^y^ s {^QDy -f rztaY, which resolves itself into the two constituents (» + ^ab) xy + rz» » 0, which represent two quadrics. Digitized byVjOOQlC 21 2779* (Proposed by Profeseor STi.yx8TB]|.)-— Sbow that the fkmotui equation between arc, angle of contingence, and perpendicular on tangent to a correi viz., p -¥ ^ »> — , may be interpreted as affirming the eelf-emdemt propoeiiion that the radios of car^atnre is the distance between the tangent at any point of a curve and the tangent at the corresponding pcnnt of its second evolute; and account for the connecting sign between p and -jB. in the above equation hemgplne and not minus. Solution ly J, J. Walkbb, M.A. 1. Let (r, p, fp) refer to the given curve ; and let {ri, p^ ^i), (»Vt />& ^ refer similarly to its first and second evolutes respectively. Then it is evi- dent that p +|>s = pt where p — ^ is the radius of curvature of the given curve. Now "" ^ = g. ana P,-g-$ = ^ whence p+g_p»g. 2. In what precedes, the case has been considered of p>p. But if p<p, t. e.> p—p% — pt BB p increases pi decreases, as is evident on drawing the figure; so that P2 — - 3^^ » — ^. Hence the sign between j? and J? is in both cases pUte. 2719* (Proposed by Professor Whitwobth.)— 1. The six points which, in conjunction with any common transversal, divide harmonically the six sides of any tetrastigm, lie all on one eonic, which also passes through the three points of intersection of opposite sides of the tetrastigm. 2. Of the six points, if those pairs which lie on opposite sides of the tetrastigm be joined by straight lines, the three straight lines thus drawn are concurrent, and their point of intersection is the pole of the transversal with respect to the conic. I. Solution by W.S.MoCay,B.JL It is not hard to prove this directly by Pascal's theorem ; but it is easier to consider the line at infinity, then we know (1) that the locus of centres of conies through the four points is a conic through the six middle points of the lines of the system, and that three other points on the locus are the three intersections of opposite lines of the system, and (2) that the centre of this conic is the intersection of the three Unes joining the middle points of opposite lines ; for these three lines meet and bislBct each other in the centre of gpravity <^ equal weights at the four vertices of the tetrastigm. Digitized byVjOOQlC 22 II. SohHon hy F. D. Thousov, M.A. 1. This follows at once from tho well known theorem, that if a system of oonics he described about a given qaadrilateral, the locos of the pole of a fixed line with respect to these conies is another conic. This theorem may be proved analytically, as in Salmon ; or geometrically, as follows : The given transversal is cut in points in involution by any conic of the system, and by the opposite sides of uie quadrilateral ; and the double points of the involution are the points where two conies of the system touok the transversal. Now, for each of the points where the transversal cuts the required locus, the pole lies on its )poIar, and therefore must be the point of contact of the transversal with a conic of the system. Hence the only points in which the transversal meets the required locus are the two double points of the involution determined by the intersections of the transversal and the opposite sides and diagonals of the quadrangle. Hence the required locus is a conic section, and it is easily seen to pass through the points mentioned in the Question. 2. The former part of this proposition may be derived by projection from well known properties of the middle points of the ades of a quadbUateral ; or directly, as follows : Let the transversal meet the sides in P, Q, B, S, T, V; and let the points of harmonic section be p, q, r, s, t, v. Then {B;?AP} = {D*AS}; therefore T{BpAP] « T{D»AS} -T{B*AP}; therefore Tps is a straight line. Similarly, Tqr, yrs, Yqpt B.8V, Qpv, &c. are straight lines. Hence, by the harmonic properties of a quadrilateral, pr and sq intersect in the pole of TV with respect to any conic through pqrs, and therefore the intersection of pr and sq is the pole of the given transversal with regard to the conic of the Question. Similarly, the interaection of sq and vt is the pole of PR, ». e. of the transversal, with respect to the same conic. Hence sq, pr, vt meet in the pole of the transversal. 2766. (Proposed by S. Watson.) — Three points move from a fixed point in three different but given directions, with given uniform but different velocities. Find the locus of the centre of the circle passing through them. I. Quaternion Solution by W. H. Lavbbtt, B.A. Let a, jB, 7 be the unit-vectors in the directions of motion, and at the end of any time let ^^jo, kx^fi, kx^ be the vectors to the moving points.; so that, p being the vector to th^ centre of the circle, we have (Ara-ia-rt^ - (kx^- p^ = {kx^-p^. Digitized byVjOOQlC 23 therefore l:xi^+2xiSap » Jtaff+2XiS^ « 2rj^ + 2a^Syp; hence, eliminatiiig Jk, and making obvious sab- Btitations, we get S(aa + bfi + cy)p = 0, where a+^+c » 0; and therefore the locus is a straight line perpendicular to the vector f/^w(aa+hfi-¥cy). If AB, BC, DO be the directions respectively of a,fi,yi and if, arithmetically, AB + BO » CD ; then AD is the du-ection of/. n. Quaiemhn SoUdion by F. D. Thomsoh, M.A.; and G. A. OaiLYiE. It is easy to see that the locus must be a stnught line through the origin, since the triangle formed by the moving points always remains parallel and similar to its initial position ; but it may be interesting tojgive an investi- gation of the locus by the method of Quatemiona, Let O be the point of departure ; A, B, C the points at the time t; u,v,w vectors representing the constant velocities of the Stints. Then, writing a, fi, y for OA, B, 0C« we have « _ ^ ^ 7 . ^^ ^ Now, let P be the centre of the circle circumscribing ABC, and let OP=p. Then we have to express the fiicts, first, that P is in the plane ABC» and second, that it is eqmdistant from ABO. The first condition is equivalent to the assertion that PA, AB, BC are in one pkne, or that S.PA.AB. BC = 0, or 8.(o-p)(/3-o)(7-/3) -0, or S.p(Voi3 + Vi87 + V7o)«So37, or S.p(Vtto + Vtw+ViwO«<S«iw...(l). The second condition gives (a-p)« - {$-py = (7-pft or o?-2Spa - fiP-2Spfi = 7*-2Sf^. or ' 2Sp(iB-a) » iB«-o», and 2Sp(7-/3) = 'f-fiP; or 2Sp(o-fi) « t (©»-«'), and 2Sp(wr-w) - <(ftr»-i»=) (2, 8). Eliminating t^ 8.p(yuv + Yvw + yum) _ 2Sp{v'-u) 2Sp(io-p) Suvw " t>8-i«« " «»-©» ' a system equivalent to two equations, each of the form SpSsO, and therefore representing two planes through the origin, their intersection being the locus required. Similarly, it may be shown that the locusof the intersection of perpen- diculars of the triangle ABO has for its equations [The three directions of motion are, in the first solution, supposed to be in one plane, but in the second solution, anywhere in space.] Digitized byVjOOQlC 24 274S. (Proposed by the Editos.)— Show that the average area of all the ellipeea that can be inacribed rpimetrically in a portion of a parabola cut off by a doable ordinate perpendicular to the axis^ in parts of the area of the parabola, is -^ir if (1) the centres be eqaably distributed over all possible positions on the axis, and 4^x if (2) the intersectiona of the chords of contact with the axis be eqaably distributed on the axis. I. Solution hf Stephbk Watsoit. Let AiB'^the doable ordinate, CD the axis, PQ the chOTd. of contact cutting CD in I, O the centre of the inscribed ellipse, and EG a common tangent at Q, meeting CD produced in E, and the semi-diameter O^ perpendicular to CD- produced, in 0. Put CD - a, CB = J, CO - o, OP « /3, 01 » ox ; then we have i^«OE.01-=(2lD + OI)OI-2(a-«)aa:-a«ar«, therefore a^a[{2x)^^x}, and EI(-.2DI) : IQ » EC : 0G> 2D1 " 2a " 2a3jr * therefore /i»-OQ.IQ«- .(A). '-M'-fi)'!' hence the area of the ellipse hi parts of that of the parabola it g = f{(2,)*-.}(l-(|)»} (1). In this case the number of posdble positions of O is |a, and the limits of # are and \ ; hence the re^uirad average is lf^(x)a. . ^f{i2.)^-.} {x-(|)»} {(2.)-»-i}«». . ^ ,. (2). In this case the number of positions of I is a ; and the limits of jr, and \ I hence the required average is i/-(A).(,.<») - ^f {(a,)*-,} {i-(|)»}(2,)-».i« = g,. 11. Solution by the '^BX^BOSEA, Let the aquation to the parabola be y^«4ma? (1); then, (a, /3) being the semi-axes of an inscribed ellipse, its equation will be (frfli^\^.i (2). Digitized by Google 25 Eliminating y fh>m (1) and ^2), and applying the condition for equal roots to the resultant, we find that the ellipse (2) will have doable symmetric con- tact with the given parabola (1) if iB«+2ma-5/3, or a - 5^1^ (3). 2fli Otherwise : The equation of a conic which has double contact with the parabola (1), along the chord PQ or x—z^O where z « DI, is 5f«-4ma? = it(a?-«)' (4)5 and determining Jk so that (4) may touch the double ordinate AB, we find k - ^^ ^ hence the equation of an ellipse inscribed in the parabola becomes anditssemi-axesare a^^z£, ^^^(a + ^) (g). 2a o From (3) and (6) we see that the area of the ellipse, in parts of the parabola, is gj(i/3«-i8'). or ^(a+»)(a>-»«) (7.8). In (1), a varies uniformly from to ^ ; hence, by (7), the average area of the inscribed ellipses is 5^3/* *"(ft/3-i8») d$ » ^y * (&/3-i8») (6-2/3) d» i In (2), z varies uniformly from to a; hence, by (8), the average area of the inscribed ellipses is From (7) or (8) we find that the inscribed ellipse is a maximum when ziB^a, a»^t fi — if^i its area being )ir. 2640. (Proposed by the Rev. J. Wolstenholme, M.A.)— 1. If a conic of given eccentricity e circumscribe a triangle ABC, the locus of its centre is (yz sin' A + ga? sin» B -h a?y sin' C)' ^ (g»- 2)* , a?y«(af+y + ») sin' A sin'Bsin'C 1— «' ' TOL. XI. Digitized byVjOOQlC 26 nnd (2) if it be inicribed in the triangle, the locus of the centre is . (a»8in'A-f...-f2ygginB8inCco8A-f ...y ^ 4(e»~2)\ xyz (a? +y + 2) sin* A sin'' B sin* C 1 — c' ' the triangle of reference having its angular points at the middle points of the sides of the triangle ABC. 8. Hence show that the locos of points with which as centre two similar conies can be drawn, one inscribed and the other circumscribed to the tri- angle, is two circles, one the circumscribed circle of the triangle, and the other that which has the centre of gravity and the centre of peipeudicnlars of the triangle as the ends of a diameter. I. SolvHon hff W. S. M*Cat, B.A., and the I^oposeb. 1. If 2 and A be the tangential equations of a conic and the circular points at infinity; then the values of Jfl, for which 2 + Ar^A » represents points, are proportional to the squares of the axes of the conic, and are given by the discriminating equation A' + Ae'F + eit^^O, for this is evidently so in the case a^K^ + 6=^'— 1^ + *> (X' + /i>) — 0. Hence, easily, if ^ be the angle between the asymptotes e being positive for an ellipse and negative for an hyperbola. Now for the circumscribiug conic 2 (fsfz+ffxz+hxy) s 0, — e =/« sin«A+ ... --Tigh an B sin C..., -e'«2(/cosA+^cosB+AcosC). But /:g:h^^(^x+Y+z):( ) : ( ), smA in terms of areal coordinates of centre, /:,:*-41|,: £±|y :£±£«. sm A sm B sm C the coordinates being referred to the triangle whose vertices are the middle points of the sides of the original triangle. Houce we obtain at once the first equation given in the question. 2. Again, for the inscribed conic (Za?)* + (my)* + (n«)* « 0, e » 2lmn (I sin B sin C -I- m sin A sin C -I- n sin A sin B), e'. P + iB« + n«+2»incosA+...., and Z : f» : N = sin A (— X + Y + Z) : ( ) : ( ) » or sin A : y sin B : z sin C. Hence we obtain. the second equation given in the question. 3. Eliminating e from the equations in (I) and (2), we see that if with the same point (x, y, z) as centre there can be drawn two similar conies, one inscribed in the triangle ABC and the other circumscribed about it, the point (^, y, z) must lie on one of the two circles ar2 giui A + ... + 2y« sin B sin C cos A + ... « ±2 {yz sin' A + ,„\ If the lower sign be taken, this is the circumscribed circle of the triangle ABC; and if the upper, it is the circle in which th>3 centre of perpendiculars and the centre of gravity are ends of a diameter. This latter drcle is there- fore one of the system coaxal with the circumscribed and nine-point circles. Digitized byVjOOQlC 27 II. Solutions ly the FbofoSEB. 1. The equation of a conic circumscribing the triangle ABC, referred to AB, AC as axes of Cartesian coordinates, is of the form AX(X-c) + /iXY+j^Y(Y-6) = 0; and if e be the eccentricity, (e3--2)2 ^ 4(X + y~/iiC0sA)2 , 1-ea 8in2A(4j^X-/i2) ' also for its centre X(2X-c) + /iY « y(2r-6)+/iX = 0; but if we transfer to areal coordinates measured on . the triangle ahc (the middle points of the sides of ABC), c and therefore for the centre Xc^ vK^ __ iiho (x+y)e y{x-¥z) 2y«* and the locus of the centre, when the eccentricity e is given, is (g2_2)' {2(ir + y)sin2B+y(iP + «) sin'C — 2y«sin BsinCcosAJ^ !-«" sin'A sm^B sin^C | y« (a? + y}(ar + a) -j/2;jsj which easily reduces to the form given in the question. 2. The equation of an inscribed conic is Aft With the condition Tt - ^"1 Tv - t^ « -^ ; \h cj \k h) Kk XT 1 and for the centre - » — « __ -. ; % . k 2(1— X2^ and oombimng these, we find |.|.sa-..,.|L(f.|-|). Transforming as before to the triangle <tbe, we find x«. f*-— -. ^(*+Sf)«^(ar + «)=ar(*+y + z). {x+y)(x+z) e b But, for the eccentricity e, we have the equation sin«AJJL«^i=2^n {(i?+y)2 8in3B + (a?+z)3 8m2C-2(a?+y) {x + z) mnB sinC cosA _. +.4sinB sinC cosAyg}' 4sin2Asin2Bffln»C.y«.a?(a?+y + 2:) mm (*^M^^^A -»- . ... +2ygsinBMnCco8A+ .*..)' 4ayx (a? +y + «) sin' A sin^ B sin' C Digitized byVjOOQlC l-e» 28 Another wlation may be obUuned in the following manner :-* 1. If the triangle of reference be the triangle having its angles at the middle pointe of the ndes of the triangle ABC, the equation of a conic dr- cnmscribing ABC is Z(«+y)(«+«) + m(y + s)(y + «)+ii(« + a?)(«+y)-0, and the equation of the stnught lines through A parallel to its asymptotes is whence, if ^ be the angle between the asymptotes, tan^^ « «n*Asin»BBin»C {(m-ni-Z)»-4fim} { n sin^ fi — (m + n - 2) sin B sinC cos Af m sin^C }'* Now, if e be the eccentridty and (X, T, Z) the centre, <f=^'--4cot«^, and ?_ - _? ?_ ; l-«» ^' X(Y + Z) Y(Z+X) Z(X + Y) whence, if e be given, the locus of the centre is l-d" sin'Asin'BsinaC { YZ(Z-f X)(X4-Y)-yZ«} («»-2)« " jz (X + Y) sin«B-2YZ sinB sinC cosA+Y(Z+X) sm»C}' ^ XYZ(X+Y + Z)rin«Asin8BrinSC "~ JYZsinSA+ZXsin'B + XYsin'C}'' 2. With a nmilar notation, the equation of an inscribed conic bdng ^j^j . 16 Imn (l + m + n) sin* A sin' B sin' C { r^ sin'A + m2 Bin3B + n>sin>C + 2mji sin B8inCoosA+. .. + ...}'' X T Z and — s z s -, whence the locus of the centre of an inscribed conic of Imn g^ven eccentricity e is (gg-2)8 (XgsinlA+...-f2YZsinBMnCco8A-h...)a l-fl« " 4XYZ(X + Y + Z)smSAffln»Bsin«C ' 2781. (Proposed by the Rev, J. Wolbtbnholmb, M.A.)— Prove that the maximum value of the common chord of an ellipse and its drcle of curva- ture, a and b being the semiaxes of the ellipse, is I. Solution hif Matthew Collins, BA. Let A A' a 2a be the major axis, O the centre, and PQ'a ^ =r the tangent PQ; then PQ'R is well known to be a chord of the drcle oeculating the ellipse at P. Let OMsor, the perpendicular MP— y; then we have Digitized byVjOOQlC 29 Again, drawing FQ'B' parallel to PQ, and putting Q'B » z, we have QA.QA^ _Q^A.<yA^ Q^A.<yA^ . ^-«' ^ «'--(^~ - 2arj a' V ar / ^ (ag~a?g)(4a:»~o») 4a^-ga a?» therefore PB « < + « «^ < ^1 + 1^ - < ^l^y Now, when this is a maximum, must also he a maximum ; its differential coefficient equated to zero gives 3(a'-i2)ar*-2o«a!3(2a3-52) + a« « 0, whence — = 2a»-i3+K - 0Q« where R « (a<-a2^ + J<)* is plainly > a'-^ftS; and therefore, as 0Q2>aS, the proper value of OQ* is 2«'— J2 + r . hence »« a«^ ^ 2a«-62+R _p _^ 2g«--&«-R _ 2R + 2a»-ft8 . 8 8 ' therefore PR2« 15 <»ar*- 15^' -15^ 2R + 2«»~y a* /a*y 8 *(R + 2aS-62/ _16a* (2R + 2o»- h^ (2a^ -f &« - R)a 3 ' {3a2(a3-6«)}* "^ 27I^*^^^--®^'('^-^>^(^«'-*'^l ' 27(a'-ft3)2 {^R'" (2«^-^) [3R^-(2«»-6^)2] the required expression for the square of the maximum common chord. II. Solution hy the Proposeb. The circle of curvature of the ellipse ^ + ^ « 1, at the point whose ec- a' or centric angle is 0, meets the ellipse again in the point (a cos 36, -6 sin 30), and the square on the common chord is therefore Digitized byVjOOQlC 30 «3«48m«26(a»8m«6 + i«co8«e)«2(l-»«)(a« + ft«-A), if c = co82#. For a maximam or minimam value, either ^-0, or 8c3«'-2(a«+6=)«-c» - 0, do of which the former gives u^O; and since «' is essentially positive, this is the minimum value. The maximum value is then given by that root of the latter equation which is numerically less than 1, or by a2 + ^,2(a<~aa6a4-&<)* ^ ga.4-y~2R For this value we have i«2 « ?i J2(a5 + bO^ + c«}- ?lfl!±^^ {2 (a« + b«) « + ca} --2c2fi + 2(a2i-b«) d 3c* 9e> ' ' 8£!» 8 3 „_^«{(a' + b»)> + 3(a»-b»)'} + ^(a»+b>) , - -i?. (a«-a»l)S+M)(a« + 6S-2B)+ ^(aS+JO _ ii {2R'-(a'+6'j [(a<-a»6»+M)-3(o'-bO']} " WW^ '^^"*~ a>6' + M)5-(o»+ 6») (2a»-6») (26»-a')} . The longest common chord has therefore the value given in the question. 5. (Proposed by H. R. Gsebb, B.j^.) — Determine in trilinear co- ordinates the foot of the perpendicular firom the point (a;', y\ s^ on the line Ix + my + nz'BiOi also the reflexion of the same point with regard to the same Une. Solution hy James Dale. If the line La?+My + N««0 (1) be perpendicular to Ix-hmy -f ns » . . . . • (2), and pass through (a/, y, /), we have the conditions Lar'+My + Na'-O, W + Mm'-^Nn'^O (where i'= Z - n cos B— m cos C, with corresponding values for m', «') ; L ^ M ^ N therefore I »»' »' I \n' V [^ \ V m' \ y' z'\ \z' j/\ I a?' / Digitized byVjOOQlC 31 The intersection of (1) and (2), that is, the foot of the perpendicular from the given point on the given line, is J_ . m n M N n I N L f_ / m L M and substituting for L, M, N, we get « V C m n n' V V mf 2! sf ^ y' n I V m' m' n' a// y' ^ I m y' 7! Uaf 2a a b I m n m' n' . «' V V m' y ^ 7! a/ X' y' The coordinates of the point of reflexion (a?i,yi, «i) may now be found from the equations 2{ = a:' + x^, 2ij = y + yi, 2{ = ^ + Zi, 230L (Proposed by W. K. Clifpobd.)— A circle is drawn so that its radical axis with respect to the focus S of a parabola is a tangent to the parabola ; if a tangent to the circle cut the parabola in A, B, and if SC, bisecting the angle ASB, cut AB in C, the locus of C is a straight line. Solution by the Rev. J. Womtenholme, M.A. Through the foci S, S' of a conic draw a drcle, and from any point T on this circle draw two tangents ; the straight lines bisecting the angles between these tangents will bisect the angles between TS and TS', and will therefore pass through the fixed points where the circle meets the minor axis. Reciprocate the sys- tem on S; then to the circle corresponds a parabola having S for its focus, and to the ellipse a circle having for radical axis to it and the point-circle S a tangent to the parabola (the reciprocal of S') ; to the point T corresponds a tangent to the parabola ; to the tangents the two points A, B ; and to the straight lines bisecting the angle, points on AB where the straight lines bisecting ASB meet it. Whence the theorem. 2456. (Proposed by Dr. Shaw.)— Show that if an ellipse pass through the centre of a hyperbola, and have its foci on the hyperbola's asymptotes ; (a) the hyperbola passes through the centre of the ellipse, {b) the axes of each curve are respectively tangent and normal to each other, and (c) the two axes which are also noimals are equal to each other. Digitized by Google 32 SoUtHon by W. H. Latbbtt, B.A. (a ) Let the equation to the ellipse be — + ^ s 1 ; therefore the equa- tions to the asymptotes of the hyperbola are ay'-y(a?'-c)-c/«0, and ay'-y(»' + c) + cy'= 0, where the centre (a/, y) of the hyperbohi is on the ellipse. Hence the equa- tion to the hyperlx)la is where Xr^ is a constant. The cunre then will not pass through the centre of the ellipse unless k=^0, when the equation becomes *y»+5^2(^3_ca)_2aya:y + 2c«yy'-0 (1). (6.) If, in (1), we put yssO, we have aj'^O; therefore the major axis of the ellipse is a tangent to the hyperbola at the ori$;:in, and the m^nor axis is a normal; also the axes of (l), being the bisectors of the focal distances, are tangent and normal to the ellipse. (c.) Again, transform the equation (1) so that (a/,y) may be the origin, and the tangent and normal to the ellipse the coordinate axes ; the trai^ortned equation will be ? ^ ^ ^ = 1. Therefore the axis of the hyperbola, which is a normal to the ellipse, is equal to h, and is equal to the minor axis of the ellipse, which, by (b), is the normal to the hyperbola. 2653. (Proposed by J. Gbi7VITH8, M.A.)— 1. If the centre of one of the four circles which touch the sides of a triangle self-conjugate with respect to a parabola lies on the directrix ; show that its circumference passes through the focus. 2. If the point of intersection of the three perpendiculars of a triangle in- scribed in a parabola lies on the directrix ; show that the self-conjugate circle of this triangle passes through the focus of the curve. Solution Ity the Rev. J. Wolstenholme, M.A. 1. If we take the triangle as that of reference, the equation of the para- bola is 2a^+ fiMf^-k-ns^ » 0, with the condition - + — + ^ -> 0. I m n The equation of the directrix is x ( — + Sl\ + .... +....«0; \n mj which will pass through the centre of the inscribed circle, if '(t*5- + ....«0, or ^+.... + ....«0. al Hence J (1+f - ili j _ „ (1±* _ *1^) . , ^112 _ ItiV or la (A— c) « mb {c^a) ^ nc (a— b). Digitized byVjOOQlC 88 But the foeos is giren by tbe eqnataons — -r — .-.,.., or I •i6«+nc» b ^ c c—a a—h and this point manifestly lies on the inscribed circle («cot j A/ + ... + ... »0. This point is of course the point of contact of the inscribed, or escribed, circle with the nine-point circle. It may be noticed that the tangent to this circle at the focus of the parabola is the polar with respect to the parabola of the centre of the circle, its equation being -^— + -i^ + -f— » 0. h—c c^a a^b There is no difference in the work when one of the other circles is taken, beyond changing the signs of one of the three, a, 6, o. 2. This is only the reciprocal of the above proposition with respect to the focus of the paraboU. 2718. (Proposed by Professor Caylby.) — Find in piano the locus of a point P, such that from it two given points A, C, and two given points B, D, subtend equal angles. 2767. (Proposed by Professor Cayley.)— If *o'+yo" = l, la?, y, 1 I « L ; show that each of the equations ag(x-Xoy + &g(jr->.ye)« ^ a«(a?>ar,y + &a(y-yt)« ^^^^ (a?aro+5(yo-l)* (a^ari+yyi-l? g»(a?~aroy-|.6«(y-yoy ^ (^(x-Xi)^ + hHy-y{)^ /g)^ (ajyo-»oy)^-(*-'o)*-(y-yo)' (a?yi-a?iv)«- («-a?i)*-(y-yO»"* represents the right line L « and a cubic curve. 1819. (Proposed by C. Taylor, M. A.) —From two fixed points on a g^ven conic pairs of tangents are drawn to a variable confocal conic, and with the fixed points as foci a conic is described passing through any one of the four points of intersection. Show that its tangent or normal at that point passes through a fixed point. Solution of the above Problems by Pbofbssob Caylby. I. It is easy to see that drawing through the points A, C a circle, and through fi, D a circle, such that the radii of the two circles are proportional to the lengths AC, BD, then that the required locus is that of the inter- sei^tions of the two variable circles. Take AC « 2Z, MO perpendicular to it at its middle point M, and ^p; a, b the coordinates of M, and A the inclination of p to the axis of a; ; then coordinates of O are a +>» cos A, b+p sin A, coordinates of A, C BTea±l sin A, d + / cos A, VOL. XI. D Digitized byVjOOQlC 84 and hence the equation of a circle, centre O and passing through A, C, is (x-a-p C08X)'+ (jf—b—p sin X)« « P+i?*; or, what is the same thing, (»-a)«+(y-&)'-l« « 2p[(a?-a) cos X+(y-J) sin x]. If 2m, 9, 0, <{, ^ refer in like manner to the points B, D, then the equation of a drde centre say Q^ and passing through B, D, is (»-c)»+ (y-d)»-iii^'- 22 [(a:-c) cos ^ + (y-d) sin /«]. And the concUtion as to the radii is I? +p» : nfi-^f^t- : m^ that is, j^ : ^2= r^ : ffi', or p : j[ » + / : »i. And we thus have for the equation of the re- quired locus (x-a) cos X -»- Cy-6) sin X " "• m (4f-c) co8/A + (y-(0 sin /i' ▼iz., the locus is composed of two cnhics, which are at once seen to he circular cubics. One of these will however belong (at least for some posi- tions of the four points) to the esse of the subtended angles being equal, the other to that of the subtended angles being supplementary ; and we may say that the required locus is a circular cubic. 2. If two of the points coincide, suppose C, D at T; then, taking T as the origin, we may write a a { sin X, ft » —2 cos X, CB^msin^, d*- mcos/i. and the equation becomes or +y^a' 21 {x sin X— y cos X) jT cos X -f- y nn X _ , i g^+.y' •»■ 2m {x sin fu—y cos /i) " m xoos/A+ysin^ ▼iz., this is (as^+y*) [m (a? cos /i +y sin /i) + I (a? cos X +y sin X)] ^2lm{{x sin X— y cos X) {x cosfi-^y sin /i) + (x sin /A— y cos fi) {x cos X +y sin x)^ = 0. Taking the lower signs, the term in { J is (ac^+y'^) sin (X— /i), and the equa- tion is («» + y3) ^„| (ay cos /A 4-y sin /i) + Z (a? cos X +y sin x)— 22m sin (X— /*)} =0, viz., this is rr' + y' = 0, and a line which is readily seen to be the line AB ; and in iact from any point whatever of this line the points A, T and the points B, T subtend supplementary angles. Taking the upper signs, the equation is (a* +y*) [w (a? cos /i +y sin /:*)— Z (ar cos X +y sin X)] -2^{(a^^-y3) sin (x + fi)-ary co8(X + m)} = 0, which is the locus for equal angles, a circular cubic as in the case of the four distinct points. 8. The Question is connected with Question 1819, which is given above. ' In fact, taking A, B for the fixed points on the given conic, and P for the intersection of any two of the tangents, if in the conic (foci A, B) which passes Digitized by GooqIc 35 throngh P, the tangent or normal at P passes through a fixed point T, then it is clear that at P the points A , T and B, T subtend equal angles, and conse- quently the locus of P should 'he a circular cubic as above. The theorem will therefore be proved if it be shown that the locus of P considered as the intersection of tangents from A, B to the variable confocal conic is in fact the foregoing circular cubic. I remark that the fixed point T is in fact the in- tersection of the tangents AT, BT to the given conic at the points A, B respectively. 4. Consider the points A, 6, (which we may in the first instance take to be arbitrary points, but we shall afterwards suppose them to be situate on the conic — + ^ » 1,) and from each of them draw a pair of tangents to the conibcal conic \- -^L. » 1. Take {xq, yd) for the coordinates of A, and <r + h 0* + h (^ yd ^^^ those of B ; then the equation of the pair of tangents from A is or, what is the same thing, (a«tA) (6«+A)" a« + A "" b« + A ' that is, (a!yo-*oy)*-('f>*+*) («-afo)'-(»*+*)Cy-yo)*=0, or as this may also be written and nmilarly for the tangents from B we have (ayi-ari|/)a-b3(a? ^x{f-a» (y-yi)« = h [(jr-afi^a + Cy-yO*] 5 in which equations the points {xq, yo)» {^i* yi) ai^ in &ct any two points whatever. 6. Eliminating h, we have as the locus of the intersection of the tang^enta (ayo-gbyy-&' (x-xo)^-a? (y-yp)' ^ (ayi-a?iy)'-^(g-a?i)^~«'Cy-yi)' (a?-aro)« + (y-yo)' (a?-a?i)2 + (y-yi)« ' which is apparently a quartic curve ; but it is obvious d priori that the locus includes as part of itself the line AB which joins the two given points. In fiict, there is in the series of confocal conies one conic which touches the line in question, and since for this conic one of the tangents from A and also one of the tangents from B is the line AB, we see that every point of the line AB belongs to the required locus. The locus is thus made up of the line in question and of the cubic curve. 6. To effect the reduction it will be convement to write ax, "by in the place ofx,y, (outo, b^g, axi, byi in place ofx^y^ x^ yi,) and thus consider the equation under the form a«(ar-aro)»-f-l>«(y-yo)« ^ aH'-^xf-^^ {V-Vi? (ayo-'qy)*— («-a?o)'-(y-yo)* (a?yi-a:iy)^-(a?-a?i)«-(y-yi)» It is to be shown that this equation represents the line L«0, and a cubic curve. Digitized byVjOOQlC 36 Writing for a moment «o-» + 6». Vo-y + ito. andiTi -» + |, y, -y + iH, the equation becomes and hence, multiplving out, the equation is at once seen to contain the factor ^i|i— liiio (which is in fiict the determinant just mentioned), and when divested of this ikctor the equation is -b« [(y"-l) (&i?j + 4ii?o)-2aryiyoi?i]. Writing herein for ^q, %, |,, iy, their talues, and consequently &€i - *■-» (aro+a^i) +«o»i . loiJi + fii?© - 2apy-ar(yo+yi)-y (aro+a?i) + arQ5fi + fl?iyoi and arranging the terms, the equation is found to be (a«af«+6»y») ["a?(yi + yo)-!/(*i + aro)] + (a»*« + bV)(a^ayi + ^a'o) - 2ary [a« (1 + a?oari)- b» (1 + yoyO] + (a«-b«) [a?(yi + yo) + y (a?i + a?o)-(aroyi + a?iyo^] - 0, which is the required cubic curve. 7. Restoring the original coordinates, or writing -, ^, ??, &c. in place of aba *» y> ^'o, &c., we have (*' + y*) C-a^ (yi + yo) + y (a^i + aro)] + (a?»-y8)(aroyi + *iyo)-2a?y {a^-l^ + Wffti-yia\) + (a2_ba)[flj(yi + yo) + y (ari + aro)-(a?flyi + ariyo)] = 0, which is a circular cubic the locus of the intersections of the tangents from the arbitrary pomts (a?©, yo), (a?i, yO to the series of confocal conies . X* y* o^Ta "*" 6«Ta "^ ^ ' *^® origin of the coordinates is at the centre of the conies. 8. Supposing that the points (aro, yo)» (*i> yi) are on the conic ^ + I, - 1, and that we have consequently ^% ^ « 1, 5l! +^ = 1, the equations of the tangents at these points respectively are and hence, writing for shortness a=yo-yi, /3«ari-afo, 7-aroyi-a?iyo. we find * "TT'^'*"" — as the coordmates of the point of intersection T, of the 7 7 two tangents; and in order to transform to this point as origin, we must in place of », y write ar- ^, y- ^2i respectively. Or what is more con- venient, we may m the equation at the end of (6), in which it is to be now as8umedthata;e2 + yo«=.l, *|« + y,««l, write «-^ y ^ 5 for *, y. and 7 Digitized byVjOOQlC 87 then restore the original coordinates by writing -, ^, ^, &e^ for «, y, ttf^, Ac, aha and f , ?-, JL foi a, ^ y, these quantities throughout signifying a »yo— yi, h a ah fi s a^— jTo, 7 ss a?oyi— *iyo« I howerer obtained the equation referred to the point T as origin by a dUferent process, as follows:— 9. Starting from the equation at the commencement of (5), I found that tie points (xq, y©), (a^i» yi) lacing od could be transformed into the form the points (d?o>yo)> (^i>yi) being on the conic ^ ^V-^ ^J, the equation a* o* an equation wluch (not, as the original one, for all valoes of (xq, y^, (4^, ^i), but) for values of (jto, ^o), («i, yO* »«ch that f^ + ^ = l, ^ + ?l' „ 1, a^ or a* 0^ breaks up into the line AB and a cubic curve. 10. To simplify the transformation, write as before ax, by, tueo, &c., for «, y, dPo» &C. We have thus to consider the equation (a?aro+yyo-l)' («?«?! +yyi-l)' ' where xf-\-y^ » 1, x^+pi^ — I9 and which equation, I say, breaks up into the line L»0, and into a cubic. Write for shortness a = yo— ^b fi ■■ *i— fl?o, y = fl?Qyi-"«iyo» so that the equation of the last mentioned line is av + /3y+7»0. Then it may be verified that, in virtue of the relations between (xq, yo), {xi, pi), we have identically (a?-a?o)(*ari+yyi-l) + (a?-ari)(araro+yyo-l) = (aX'¥fiy + y)^^SdLEl(yx-i-a), (a?— aro)(*a?i+yyi— 1)— (a?— a?i)(asro+yyo-l) *= /8«*— oaiy— 7y-iBj and, fflmilarly, (y-yo)(a?a?i+yyi-i)+(y-yi)(*'o+yyo-i) -(«*+/3y+7)?^* (7y+/3). (y-yoXa^ari+yyi-^-Cy-yJCipafo+yyo-l) = i3ay-ay»+7x+o. 11. The equation in question may be written a'F + ft'Q » 0, where P « (a?-aro)'(a?a?i+yyi-l)'-(a?-*i)'(«»o+yyo— 1)'» Q - (y--yo)'(a?ari+yyi-l)«-(y-yi)2(ara?o+yyo-l)^ values which are given by means of the formula just obtained ; there is a common factor ax + fiif-f-y which is to be thrown out; and we have also^ as is at once verified, ^(2JL^ » xp + xi^ ^ ^^^^ iheae equal Victors may be thrown out. We thus obtain the cubic equation a»(7« + a)(/to*-cay->y-^) + 6'(7y + /3)Oay-c8^ + 7J? + o)-0. Digitized byVjOOQlC S8 This is nmplified by writing d? — ~ for d?, ^— - for y. It thne beoomei y 7 aVr [(7«-a) (/8*-ay)-7»i^] + b'y [(7^-/3) (iB*-<v) + 7*'] = Oi or, what is the same thing, thatis, 7(«'*' + *y)0to-ay) + a»[-a^«« + (a»-y)ary] 12. Eestoring f . S, fj for a?, tt^ x^ and 1?, ?!?, «> for y, yo, yi I a a a a a a writing, consequently, ^, r, X in pkce of a,fi,y, \£ a,0,y are still used a (to to denote a » yo-Vi* iB ■■ a?!— ^^o. 7 • *oyi— *iyoi t^® equation becomes + b»[-(b^/5»-7^ay + a^a^1/^] = 0, where now, as originally, ?^ + ?^ = l, ^ + 5^ — 1; viz., this is the equation, referred to the point T as origin, of the locus of the point P con- sidered as the intersection of tangents from A, B to the variable confocal conic; and it is consequently the equation whidi would be obtained as indi- cated in (8). The locus is thus a circular cubic ; the equation is identical in form with that obtained (2) for the locus of the point at which A, T and 6, T subtend equal angles, and the complete identification of the two equations may be ^ected without difficulty. 13. I may remark that M. Chasles has given (Comptes Rendus, torn. 68^ February, 1864) the theorem that the locus of the intersections of the tan- gents drawn from a fixed conic to the conies of a system (^, y) is a curve of the order 3y. The confocal conies, qud conies touching four fixed lines, are a system (0, 1) ; hence, tajcing for the fixed conic the two points A, B, we have, as a vcoy particular case, the foregoing theorem, that the locus of the intersections of the tangents drawn from two fixed points to a variable con- focal conic is a cubic curve. 2778. (Proposed by Professor STLYEdTEB.) — Suppose /a any quantity fracUonal or integral, and write also let n be a positive integer such that /i and ^+fi have not different signs ; then it has been proved elsewhere, and may be assumed to be true, that the number of changes of sign in the series is an inferior limit to the number of imaginary roots in the equation cbir" + Cia»*''^ + Ca»*"'+.... —0. Digitized byVjOOQlC 89 1. Beqirired to prore that the ahove theorem renuuns tme when we write r »(' + !) jr(ir+l) (»'+2) prorided r is a poatiTe integer, and ^c— r + 1, /i + r do not ^QiFer in sign. 2. Afloertun whether w hdng an integer ia esaentaal to the troth of the extended theorem. SolmHoM by (he Vbjofobkr, Let /r = e^ + Ci«^~*+...., Fap- ar'+^^+^c, 1 .2 1 • 2 • 3 the alternately appearing and disappearing fiictor $ being any quantity Caf^ ferent from oiuty, c an infinitesimal of like rign with c^ r any positiTe in- teger, and ft. any qoantily (fractional or integer) such that /t, and ^c-f »-|-r do not differ in sign. ^ '•^ 1.2.. ..r "^^ ^" 1.2.. ..(r+1)^* Then, by the Postulate of the question, the number of imaginary roots in "Fx is not less than the sum of the changes of sign in the two series €«5 (tf«-l)€*; (l-fl»)c?; ....; (....); a?; a«; af-ooos; ....; a*. In each of these series, the term a*, properly speaking, has attached to itself another term^ but which may be rgected as of only infinitesimal magni- tude. In the penultimate term of the upper series, the parentheris left Tacsnt (gimilarly neglecting an infinitesimal term of higher order than the one re- tained) will be — c^'^oo <^ — ^'~^ao, according as j^ is even or odd. Thus the number of changes of sign in this series is r if r is even, and r— 1 if jr is odd. Again» the roots of Vx consist of a group of infinitely gpreat and a group of finite magnitudes. The former are the roots of the equation dPfar+c^ <= 0» or, which is ultimately the same thing, oftx' + CQ'^O; the latter ofjx = 0. The number of imaginary roots of the infinite group is therefore r or r— 1 acoor^ng as r is even or odd ; t. e., is the ssme as the number of changes of sign in the upper series. Hence the number of imaginary roots belonging to fx^O cannot be less than the number of sign-changes in the lower series; but this last number will not be affected if we multiply each quantity, a^aj,a^ by the common fiictor A^O*+l)»»»>(M+y-l) ^ or, which is the same thing, if we write r + l (v+l)(i' + 2) Let now r+l*/, /A + r*/, so that |i«|i'— r -«/—/+ 1, and /i + n + r «■ f*'+ «. Then, suppressing the accents, we see that, » being any positive integer, and /I not iutermeduite between r— 1 and — «, if c^j^oo; ei^^an c j = ^i^i±i) a , ; .... Digitized byVjOOQlC 40 the number of imaginary roots infx is not less than that of ngn-changes in ''o*; o^—aQ<H* «a?— «iOs5 •••• When v»l and fi^^n^ we &11 back again npon Newton's mle. One feels strongly inclined (on more than one ground) to suspect that v need not be integer, but I do not see my way to a proof. As a corollary, it may be noticed that the equation ^»+^a;»-l+M> + l)^-2+ r r (i' + 1 ) or more generally, the real function (where 0^ does not exceed unity) V y^y + i) y{y + l){y + 2){y + d) ) \y ^y(y + l)(ir+2) / when r is a positive integer, and /a— r + 1, /i-fn have the same sign (i. e. when /A is not intermediate between r— 1 and — n), cannot have more than one real root. N.B. — It is worthy of notice, that if we make fx become the middle part of a function of x having for its roots a group of infinitely greats, a group of infinitely smalls, and the roots of fx itself, we may, by a method of de- monstration similar to that here employed, pass direct from Newton's rule (as given by Newton) to the final extension of it contained in the question above, but with the difference that, so far as the force of such demonstration takes effect, fi as well as p will have to be integral, a restriction we know to be superfluous. This is one of the reasons for doubting the necessity of i' being an integer. For the proof of the Postulate, see the Proceedings of the London Mathe* matical Society, No. II. The above demonstration would have been just as good if only the first powers of the infinitesimal €, instead of the ascending scale, had been used in <^ throughout ; but I have not thought it worth while to disturb the text by making this simpUficatioD, which the reader can . do for himself. To jmd the number of permutations of n things taken r together. By C. R. RIPPIN, M.A. Let P (n, r) denote the number. Now, if » things be arranged in all possible ways r at a time, each will stand first as often as the remaining »— 1 can be arranged r— 1 at a time,that is, P (» - 1, r— 1) times ; but the whole number of permutations is secured by making each thing stand first as ofben as possible; and therefore, since there are » things, we must have P(w,r)-«.P(n-l,r-l). Similarly P(n-l,r-l) = (n-l) . P(n-2,r-2), &C. =: &C. P(«-r + 2, 2)«(»-r + 2).P(n-r+l,l)«(n-r+2)(»-r+l.) Multiply and simplify, then we obtain P(n, r) = w(w-l)(w-2)....(«-r+2)(»-r+l). Digitized byVjOOQlC 41 2798. (Propofied by J. J. Walexb, M.A.)— To find the ocmdition that a given point (a?, y, z) may be exterior to the given conic S = (A, B, C, P, G, H jar.y, 2)' « 0. Solution hy F. D. TnoHBOir, M.A. If a point be without a conic^ its polar cats the conic in two real points. Now let \x ■\' f»y •{' vz — be the equation to polar of x^ y, z. Then, for the points of intersection of the polar with the conic, we have the equation (A, B, C, P, G, H^-(fiy+i«), Xy, \zf - 0. Expanding this as a quadratic in £, and writing down the condition for z real roots, we have (F2-BC, G«-CA, HS-AB. AP-GH, BG-PH, CH-FGjx,/u,ir)' > 0; (2S c^S c^S or, Binoe x : /i : y - ^ : ^ : ^, ax dy dz (A'.B'.C.F'.G'.H'X|. |. f)'>0 (1). where A^ B% &c. are the inverte coefficients. The condition (1) reduces to (AF« + BG«+CH«-ABC-2PGH)S>0, or AS > (2). In the particular case where |/»««0, S «» ksfi and is positive; therefore the condition becomes A > 0. 1012. (Proposed by Professor MAiTErHBiH.)— If a and I be the points of contact, with a curve of the third class, of a double tangent ; and if this tangent be intersected in 4»» n, p by the three tangents to the curve which can be drawn from any point in the plane, then f^'^'^^'f^ » ^, where hm,on,})p ph pQ^ Pb "® ^® "^ ^ curvature at the points a and b. Solution by the Rev. J. Wolbtenhouce, M.A. Take {x, y, z) for tangential coordinates, and («, $, y) for areal coordinates. The tangential equation of a curve of the third class, to which a==:0 is a double tangent at the points where it meets /3»0, 7»0, is of the form an/« + A3^ + By»s + Cyx«+D«»-0 ..............(1). For the tangents through an arbitrary point (jp, g, r) we sh%ll have px + qy-t-rz^^O (2) , VOL. XI. E Digitized byVjOOQlC 42 wbicb» oomlnuod with (1), leadi to the eobie i>(Ay» + By»t + Cy«*+D«>)-y«{5^+r2) -O; to tbat }(lSl^ « 5 and b independent of (p , q, r). But the tangent «l2,Z, A x^a+ififi + Xiy^O meets the straight line a ■ in the pcnnt where ,;,+,.,-0. or f?-«;, whence ^^ l.«».t«.t. To inye^-tigate this constant, take a point P near a, from whidi draw three tangent?, and let Z Psm » 8^, I^ » 8^\ Then the constant re- qnired is the Umit of ^-'^^'^ when M moTCB np to h, and n and |? to a, - hunt of - — - — J21. « ' limit of ^*-°P; -A lunitof ^*-^P; Pi, ^M' and making P approach its limit in such a cUrection that an^^ap^ we shaU therefore have ultimately — ^ -> 1, and ap » ip^^^' a -^. an p^ [A solution by Professor Cbexoha has been given on p. 88 of Vol. YII. of the Meprint,'] and ^^^^± S^ 8^' 8^' 2740. (PlDposed by W. S. BirBNSiDS, M.A.)~Prove that the envdope of a line given by an equation of the form Loos2^ + Mnn2^ + Pcos^ + Qnn^ + R»0 may be obtained as the discriminant of a cubic equation, and form this equation. Solution hy the PR0F08EB. The above equation is plainly reducible to the form {abcfgh) (cos^,sin^,l)> « 0, which ftirther is equivalent to (abefyh) ft, i|, 0' " or U«0, with C + ij*— f — or V=sO; whence, if we find the condition that the conies U and y touch (conndering a b efg A as constants), we shall have the equa- tion of the envelope required. And this is most easily done by forming the discriminant of U + XV » 0, w. (ABCD) (1, X)*, and then equating to zero the discriminant of this cubic. Digitized by Google 43 263d. (Propoeed by Professor Stltbstsb.)— 1. If A, B be two confbcal conies, and A' be a third oonic having double contact with A, show that there exists a fourth oonic B' having doable con- tact with B and confocal with A'. 2. If A, B be two confocal qoadrics, and A' be a third quadric having continuous contact with A, show that there exists a fourth quadric B' having continuous contact with B and confocal with A'. SoUtHon h^ the Rev. J. Wolstbkholub, M. A. 1. Generalized by projection, this amounts to : Given two conies A, B inscribed in the same quadrilateral (of which 00' is a diagonal) ; there can be described two conies A', B' having double contact with A, B respectively, and such that the tangents to the two from O, O' are the same ; or (to ex- press it more clearly) two of whose common tangents intersect in O and the other two in O'. Beciprocating this, we have two conies a, b passing through four points lying on the straight lines o, o\ and the theorem to be proved is that two conies a\ h* can be formed having double contact with a, h respectively and meeting the straight lines o, cf in the same points. The simplest case to which this general theorem can be reduced by projection appears to be: Given two conies A, B similar and umilarly situated, two circles can be drawn having double contiict with A, B respectively, and meet- ing their common chord in the same two points. This is not at all hard to prove; but as I have not seen how to get a neat solution, I refrHin. (Proposed by H. B. Gbbeb, B.A.) — Given a cubic curve K, and a ix>iut on it p ; through P is drawn any transversal meeting K attain in m and », and on it is taken a point x such that the anharmonic ratio (jpxmn) shnll be equal to a given quantity. Prove that the locus of x is a quartic curve with a point of osculation at j7, touching E at ;? (counting for four points of intersection) and in four other points. SoUdion hy James Dalb. Beferring E to rectangular axes passing through p, its equation is of the form Aaj»+Ba?^+CayS+iy + E«« + Pay-t.Gy2 + Ha? + Ey = 0, which, being transformed to polar coordinates, becomes r»(A cos? a + B cos* 6 sin 6 + C cos sin* + D sin» 0) + r(Eco8»a + Pcos^sina + Gsiu« «) + Hcosa + Esina «= 0. Let now a? be a point on any radius vector cutting the curve in m and ii, so that the anharmonic ratio » 2 : m; then, putting p«=r, />»i=ri, pa^rj, we get r {(Z-*») ('•i + r,) + (i + fii)(ri-r8)J « 2 (i-m)rir2. Substituting the values (ri + ra), (n— •'s)* •'a^'a* obtained from the equation to the curve, and returning to x and y coordinates, we get Digitized byVjOOQlC 44 o the equation to the locns of x, which is eyidestly of the fourth degree. Let Edr* + Fiy + Gy* + 2 (Hd? + Ey) « 0, which is the eqnalaon of the polar conic of p with respect to E, he represented hy P^; and let Ha?+ Ky » 0, which is the common taneent at pi to P and K, he represented by T|, and the cuhic by Eg; then the locns of » may he written «mP,«-.2(Z+m)9Ti.E3; and it may he shown that this cnnre touches Eg in the points common to T] and P2, that is^ in two coincident points at the origin, T| being the common tangent to the three cnrveSf and in fonr other points. 2494. (Proposed by Professor Etbbstt.)— On each side of a hexagon as base, a triangle is described by prodncing the two adjacent sides to meet, and a second hexagon is formed by joining the vertices of these triangles in order. Show that if either of these two hexagons can be inscribed or circnmscribed to a conic, the other can be circumscribed or inscribed to a conic. 2513. (Proposed by Professor Eyibett.)— Let A, B, C, D, E, F he six points. Let BF, CD meet in a ; AB, CE in i ; AB, CD in c ; AF, CE in d; BF, DE in ; and AF, DE in/. Show that, if either of the hexagons •ABCDEF, abcdrf can be inscribed or circnmscribed to a conic, the other can be circumscribed or inscribed to a conic. (Proposed by Professor Eyebett.)— If three conies cut one another so that every two of them have a common chord which is also a chord of a fourth conic^ the other three chords of intersection of the three conies meet in a point. Solution by the Pbofoseb. Theobeu I. [Quest. 2539] Let S be the fourth conic (». e,, let its equation be SssO), and a, jS, 7 the three chords common to it and to the others two by two; then S— /JB7, S— wyo, S— najS will be the three conies. By sub- traction between the two last, we have mya—na$ or a [ ^ — 7V denoting a pair of lines which must be chords of intersection of two of the three conies; and it is obvious that _-.!, ~ — -» -— — meet in a point. m n H II m Theobeu II. If the three conies break up into pairs of lines, the foregoing theorem has in general, for any given values of a, iS, 7, eight distinct appli- cations, that is to say^ there will be eight points in each of which three lines meet. There are two values of I which make S— l^ break up into a pair of lines. Digitized by GooqIc 45 For instanoe, if AC and BD are /3and>, 8—^ may denote ather AB and DO or AD and BC. We have thus ax piurs of lines which we may denote by 8 — Ifiy, 8 — mya, S - najS, 8 - r/iy, 8 - m'-yo, S - n'a^ and eight points of meeting, yiz., TLQc 1. ^^l^y. «»A«t; "^^^H, &c, which we may denote by Imn, Vm'n', Vmn, Im'n^ Imnf, lmfu\ Vmn\ Vw!n ; and the piurs of lines may in like manner be denoted by I, l\ m, m\ n, »'• Each of the lines passes throngh two of the points. Let A, B, C, D, £, F be any nx points on a conic, and let a, ^, 7 denote AD, BE, OF respectively; then ; is BC and £F, m is CD and FA, 11 is DE and AB; r is BF and CE, nl is FD and AC, n' is BD and £A. J. — 2 will denote the line joining the intersection of AB and CD to that of m % FA and DE, which we shall indicate by the notation ''AB, CD to FA, DE." The eight sets of lines are AB, CD to PA, DE ; BC, DE to AB, EF 5 CD, EF to BC, FA^ AC, BD to DP, EA J BD, CE to EA, PB ; CE, DP to FB, AC AB, CD to PA, DE 5 AB, CE to DE, BP; AP, CE to CD, BP BC, DE to AB, EF; BC, DP to EF, CA ; BA, DP to DE, CA CD, EF to BC, FA; CD, EA to FA, DB ; CB, EA to EF, DB AC, BD to DP, EA ; EF, BD to BC, EA; EF, AC to BC, DP BD, CE to EA, FB ; PA, CE to CD, PB; PA, BD to CD, EA CE, DP to FB, AC; AB, DP to DE, AC ; AB, CE to DE, PB 5 Vm'n Vmn' Im'n Imn* Im'n Vmnf Vm'n The figares of the first two cases are subjoined. In both of them, as here lettered, the lines meeting in a point are OL, HM, and EX. It will be observed that the hexagon OHKLMN is formed from ABCDEF, in Fig. 1, in the same way as ABCDEP is formed from GHELMN, in Fig. 2. A similar relation con- nects ^mn with toV, ^m'nwith Vnai^ and ftim' with Vm'n. The following figure, in which the conic may 1m supposed to pass either through A, B, C, D, £, F, or through a, 6, c, (f, e,/, applies to the two cases Vmn and lm'n\ the capital letters being applicable to the former and the small letters to the latter. In the former case, <u2, 60, cf will meet in a point ; in the latter, AD, BE, CF. We have thus proved that, if either of the two hexagons (mentioned in Questions 2494, 2513) can be inscribed in a conic, the diagonals of the other meet in a point ; whence it follows, by the converse of Brianchou's theorem, that this other hexagon can be drcum- scribed about a conic. na. 2. Digitized by Google 46 JITofe.— Each of the eight sets of lines is connected with a Pascars line hy the relation that, if two triangles be snch that the lines joining corresponding vertices meet in a point, the three intersections of corresponding sides will be in the same straight line. In Fig. 2 the two triangles are GKM, LNH. It will be found that the line E^l is the Pascal ABEDCF, and that £^1 m M m n is the Pascal DBEACF. 210L (Proposed by M. DAEBorx.)— Tronver los conditions n^cessaire* et snffisantes pour que les qnatre racines d'nne Equation du quatri^me degr^ forment nn qoadrilatdre inscriptible. Tronver la surface et le rayon de ce qnadrilatdre. Solution hy J. J. Wauosb, M.A. If (a, 6, c, <{) be the fonr real and podtive roots of a biquadratic s^^px^ + 90^— rd?+« s 0, it, is evident that the condition of its being possible to oonstmct a quadrilateral figure the four sides of which shall be represented in length by these roots is, simply, that the greatest of these values should be less than the sum of the other three; t. e.^ that the ezpresnou u — (a + 6 + o— d)(a + J— o + ci)(a— & + c + <i)(— a + J + c + d) should be poative. In terms of the coefficients of the biquadratic tt = 22a»i«-2a* + 8aicd = 43a»i^-(5a9)« + 8aJcrf = 4(2»-2pr+2f)-(|^-25)a+8* -i?«(4ff-;^ + 8(2*-pr). This condition being ftdfiUedy let a, & be two adjacent sides of the figure, B the angle between them, and 8 the length of the diagonal joiniug their ends; and let ^ be the angle between the remaining Bides c, d. Then, if the figure is to be inscribable in a circle of radius B, we have coge + cose'=0, whence JP = (!!i±^4±M±^!)£*; and ab-^cd cos a - * 11*--^LJl, whence an^ B = Ji L — L .^ Li 2(aA + c<f) 4(aJ + cd)« ^ 4(a& + cdP 4(ai + cd)2 The area of the quadrilateral wm\{ab-k-cd) sin B -j{p2(4^-p«) + 8(2*-pr)} (2). »3 ((a2 + 62)cd+(c2 + dS)flf6] (ah-k-cd) This numerator is a symmetric function of a, &, e, J> being equal to :i(^hcd + :ia^lrf^ « (p' - 2g) # + r» - 2gs « (jp^—Aq) « + r^, which is, plainly, essentially positive. Hence we obtain Ra_ (p'-4g)> + r> (3). Digitized byVjOOQlC 47 In the case of a convex quadrilateral, then, the only condition to be ful- filled by the real and positive roots of the biquadratic is «=0; and snch a figure is always inscribable in a real circle by giving its angles the values determined by sin ■■ —r^— — r. 2(ffJ + cd) If the quadrilateral is to be skew or twisted, jj^ (g8+y)c<g-(ca-fd»)g5 cd—ab * and that this should be positive, {(a^+ 6^ cci— (c'+ d*) oi] {fsd^ab) must be positive; or Sa^^^c*— Ja'Jcc?, i.e., r*-25*— (/i'— 2^)«, or r*— p*#, must be positive. If this condition and «»0 are satisfied, it will be found that 2616. (Proposed by C. W. MebeitibIiD, F.R.S.) — Let two intersecting tetrahedra have all their edges bisected by the same system of Cartesian axes, each axis through two opposite edges of each tetrahedron; them the solid about the origin has the origin for its centre of figure. Bohaion by the Pboposxb. The origin is of course the centre of figure of each tetrahedron separately. Moreover each tetrahedron is completely defined by the angles betwem tiie axes, and the intercepts of the edges, except that it may be inverted. Calling the intercepts a, b, o, and writing the equation ^-.-^^f.., 1 in P q r the abbreviated form pqr^ the equations of the sides of one tetrahedron are abCf a be, a be, a bo, while those of the inverse tetrahedron are abe, a be, a bo, a bo. The two systems of planes form an octohedron. I beg^n with the clearest case of intersection, in which the comers of the inner tetrahedron simply project through t^e rides of the outer one, without crossing its edges. Let the planes of the tetrahedra be represented by the following groups of planes, opposite planes (not necessarily parallel) written under one another, abe, abe, a be, a be, Imn, Imn, Imn, Imn. The secondary pyramid made by the plane Imn with the three planea abe, abe, abe has its four points (x^t/QZ^, &c. determined by the four systems of equations Digitized byVjOOQlC 48 abef imn 1 l««) tmnl Zbe[ aici 56«[ ibe\ abc) 5Jc) ubV ale) which give the folkywing Tallies:^ «o - -a. y»- -6. «»--«'. 9i - -a, fi- fcwi(a-D em»(a— n aln{b^m) 9t' ->, . _cft.(J-») "'miau+ctf _ flZm(c-ii) • (,am+bd' '*"j.(a«i + 6l)' #*- H' -e. ' The other leoondary pyramids are obtained from the first bv changing simxil- taneonaly the signs or any two of a ft o and of the oorresponding pair of { m »• Now, changing the ngns o€ ablm and leaving the signs of en unchanged is equivalent to changing the sig^ of w and y and retuning those of t and of all the constants, and so with regard to the other two permutations. Hence it is dear that the sum of the sixteen x*s is identically equal to zero, and so of the ys and s's. It follows, then, that if these nzteen points be considered o the centres of spheres of equal weight, the centre of gravity is at the origin. If, therefore, the four secondai^ pyramids have all the same volume, the proposition will be proved. Conrnder the hexahedron (or parallelepiped) aOO, ObO, 00c, aOO, 06 0, 00c, the twelve diagonals of the foces of which are the edges of the two tetrahedra (a b c) and (a b c). Let tu suppose that its alternate oomen are cut off by the four planes Imii, lam, Imn, Imn, These planes will cut off equal lengths upon the corresponding edges of the hexa- hedron, and therefore the pyramids cut off will be of equal volume^ as may eanly be shown. Next, if we consider each plane of the type (/ m a) as the common base of two pyramids, one cut off from the hexahedron, and the other from the original tetrahedron (abe), we see that these have the same altitude, and that the proportion oetween their bases is the same at all the four comen. Hence the pyramids cut off from the tetrahedron are also all equal inter te. But this bein^ the case, it has already been shown that their centre of figure is at the origin. Hence the centre of figure of the solid about the origin, that is, of the remainder of the tetrahedron (a b c), is also at the oriffin. I have chosen the simplest case for demonstration. The general proposition may be inferred from the principle of the permanence of eouivalent formfl^ or, with a little more trouble in picking out the details, may oe proved inde- pendently. Cob. 1. — ^The solid formed about the origin by any number of such inter- secting tetrahedra, and by hexahedra of the same system, has also the origin for its centre of figure. Cob. 2. — Cleavage parallel to one of the coordinate planes simply shifts the centre of figure along the corresponding coordinate axis. Cob. 8.-— If the solid be made up of any number of tetrahedra and hexa- hedra about the same axes and having their oomera on the same four dia* Digitized byVjOOQlC 49 gonals, then the effect of deavage in a tetrabedral plane is amply to shift the centre of figure along the diagonal corresponding to the cleavage plane. If there are to be no hollow spaces, there mnst of coarse be at most only one hexahedron, and one direct and one inverse tetrahedron. Octohedral cleavage is thus seen to be more complex than bexahedral, as might be expected. NoTB OTSf Question 2740: by Fbofesbob Catlby. The envelope of the curve A cos20+Bsin20 + Cco8 9 + Dsin0 + E»O, (where A, B, C, D, E are any fanctions of the coordinates, and is a variable parameter,) is obtained in the particular case E^O (Salmon's Higher Plane Curves, p. 116), and the same process is applicable in the general case where E is not ss 0. From the great variety of the problems which depend upon the determination of such an envelope, the result is an important one, and ought to befamiUarly knoum to students of analytical geometry. We have only to write 2 » cos + 1 sin 0, the trigonometrical functions are then given as mtional functions of z, and the equation is converted into a quartic equation in 2r ; the result is therefore obtained by equating to zero the dis- criminant of a quartic function. The equation, in fact, becomes that is, A(«* + l)-B»(«*-l) + C(«» + «)-Dt(»»-«) + 2Eaa = 0; or, multiplying by 12 to avoid fractions, this is (<i,J,c.<f.«)(«,l)*-0, where a->12(A-Bi), i»8(0-Dt), c»4E, « = 12(A + B0, d«8(C + Dt); and substituting in (ae-46ci + 3c*)»-27(actf-ad»-68e + 26cd-c»)2 -. 0, the equation divides by 1728, and the final result is {12(A2 + B2)-8(C2+D2) + 4E3]' - {27A ((?- D«) + 54BCD- [72 (A« + B*) + 9 (C? + D^)] E + 8E»] « - 0. It is to be noticed, that in developing the equation according to the powers of E, the terms in E^, E^ each disappear, so that the highest power is E^; the degree in the coordinates, or order of the curve, is on this account some- times lower than it wovHd otherwise have been. 2823 (Proposed by Professor Sylvesteb.)— Show that on a chess-board the chance of a rook moving from one square to another without changing TOL. XI. P Digitized byVjOOQlC 50 colour IB f; but that without altering^ the equality of the number of the black and white squares, bat only the manner of their dbtribntion, the chance may be made eqnal to J. I. Solution by Mosgan Jsinci^s, M.A. Wherever the rook may be placed on the board, there are in each of the two lateral directions in which the piece can move 3 squares of the same and 4 of a different colour to that of the square on which the rook stands. The required chance is therefore f . No possible distribution of the colours can be made which shall raise the chance to i wherever the rook may be placed ; for if or be the number of black squares in a row, then oat of the 14 squares to which the rook may move 7 squares must be of the same and 7 of different colour to that of the square occupied by the rook. Hence, placing the rook on any of the x black squares, we should have x—1 other black squares in the same row, and there- fore 8— J* other black squares in the same column; and placing the rook on any of the S—x white squares, we should have x squares in the same row, and therefore 7—x black squares in the same column, whence follows the equation ' x (9-ar) + (8-ar) (7-ar) =82 (1). The only solution of this is rr— 4; which shows that there could be only 4 black squares in any row, and in like manner in any column ; but this is inconsistent with the supposition in (1), that if there are ^ or 4 black squares in a row, there will be 9— j; or 5 in some of the columns. Hence the chance cannot be ^ for every position of the rook ; but by sup- posing that every position of the rook is equally probable, we may, by various distributions of the colours, make the total chance for all positions of the rook i. Let Xi,X29 ,... Xg be the number of black squares in the rows, yi, y2» • • • • ys » » »> columns ; then X1 + X2+..,, +xs^S2, yi-»-y2+ •• •• +^8 «« 32 (2,3); and also the following equation is easily obtainable, a^i(a?i-l) + arj(ar8-l)+....+(8-ar0(7-a?,)+.... > ^32x14, +yi (yi - 1) +ys (y2-i) +....+ (8-5^1) (7-y,) +....) which reduces to Xi^+ +yi^+.. .. = 16x18 (4). It is easily seen that, if Xi, d?2* • • • • i yi* ^3* • • • • ^ ^ ^^^ ^^ solutions, then 8— *!, 8— d:2» . • • . f 8— yi, 8—^2, .... is another set, which may be called a complementary set. To facilitate the finding of solutions of (2), (3), and (4), I have made the cueumption (involving possibly a loss of generality) that xi^+x<?+ .... +a'8^ = 16x9 = y,«+y2'+.... +^8* (6,6). Supposing that there are a rows containing 1 black square, b rows containing 2 black squares, and so on, we have a + 6 + c+...*=8, a + 26 + 3c+...«32, a + 45 + 9c + ...= 144 .... (7,8,9); whence we obtain 6 + 3c + 6rf + 10cf ... = 66, c + 3<i + 6c+... =32 ....(10,11). Digitized byVjOOQlC 51 The following sets of values of Xi, X2, . . . . ^s ^^^^ ^ found, I think, to be all which can be obtained from the positive integral solutions of (7), (10), and (11). [6, 6, 4, 4, 4, 4, 2, 2] self-complementary [6, 6, 5, 4, 3, 3, 3, 2] [6, 5, 5, 5, 4, 3, 2, 2] [7, 5, 4, 4, 4, 3, 3, 2] [6, 5, 5, 4, 4, 4, 3, 1] [7, 5, 5, 3, 3, 3, 3, 3] [5, 5, 5, 5, 5, 3, 3, 1]. The following squares are specimens of distributions obtainable from these sets of values: — 4 w w W W B B B B 4 w w W W B B B B 2 w w W W B B W W 2 w w W W B B W W 4 B B B B W W W W 4 B B B B W W W W 6 B B B B B B W W 6 B B B B B B W W 2 w W W W W B W B 4 B W W W W B B B 3 B W W W W "W B B 3 B B B W W W W W 3 B B B W W W W W 5 B B B B B W W W 6 B B B B B W B W 6 B B B B B B W W 44446622 75533333 [Professor Stlvksteb remarks that any anallagmatic distribution of the black and white squares will satisfy the condition. For, in such a square, whether the motion is from a row to a row or from a column to a column, there will by its definition be as many passages from like colour to like as firom a colour to its opposite. Suppose an anallagmatic square is made iso- chromatic, i. e., having always the same difference (positive or negative) be- tween the black and white colours in each row and column (see the diagram at the end of Vol. X. of the Reprint). Then, putting n for the number of squares in a side, and x,n—x for the number of the separate colours respec- tively, it is clear from the above solution that we must have 2ar(»-*) = J2».^zl, .-. x{n-x) » !^Zf, i.e., x , n±^n which proves that an anallagmatic square cannot be made isochromatic unless the number of places it contains is a perfect fourth power. This disposition springs directly from an improved form given to Newton's complete rule for limiting the total number of imaginary roots in an algebraic equation.] II. Solution by Matthew Colldts, B.A. 1. As each line and column of a chessboard contains 4 white and 4 black squares, therefore upon whatever square a rook be placed there are 3 other squares of the same colour, and 7 other squares altogether in the line or column in which the rook can move, and hence his chance of moving into a square of the same colour as that be left is plainly « f . 2. If the Une bisecting two opposite sides of the chessboard be the boundary between the white and the black squares, it is plain the square common to any line and column has, upon that line and column taken jointly, 10 other squares of the same colour as itself; and as there are 14 other squares alto- gether upon that line and column, hence in this case the chance of a rook moving into a square of the same colour as that it left is » -^ = f . Digitized by Google 52 8. If the diagonal dividea the white from the black aqoares, and if the aqnares through which it paaaes be white and black alternately, the required chance of a rook not changing colour when moved would plainly be 14.7 + 10,9 + 6.11+2.13 _ 6 82.14 8* 4. But if the four white squares on the diagonal occupy half of it, the other half running through four black squares, then the required chance is •^ (4.6 + 6.7 + 6.8 + 5.9 + 5.10+8.11+2.12 + 1.13) r=???. 82 . 14 ^ '448 6. Lastly, if one row consist of eight white squares, and another row of eight black squares, the remaining six rows of the chess-board being left unchanged, the required chance is then plainly 8.10+24.6 _1 82.14 "2 274L (Proposed by S. Tebat, B.A.) — A moveable event (depcpduig on the moon) happened in the year y ; show that it will happen again in the year ^ + 19(28^— 5a +86) ; where a may be 0,1,2,3; b the number of completed centuries since the year y which are not leap years; and t any arbitrary integer. Solution hy ike PaoFOSEB. Let T B f^ + 4»-f-a be the year ; then the number of days in tins interval is 865(4n + a) + fi-A«7(208» + 62a) + 5fi + a-6; therefore Sn + a— h must be diviMble by 7 ; therefore ii-»7m— 3a+d&, and 4it + a » 28m— lla+125, which must be divisible by 19 ; therefore m » 19^-3a + 5i, .'. 4fi+a » 19 (28<-5a + 86), and Y = v + 19(28/-5a + 8ft). JExan^le : Easter-day cannot happen later than April 25th ; the last time this took place was in the year 1734 ; when will it happen again? Let t=^0, assO, isl; therefore T — 1886, the year required. 2598. riVoposed by G. M. Suith, BJ^.)~A uniform rod (mass » /u) is placed inside a spherical shell (mass » m, radius = a) which, under the in- fluence of g^vity, rolls down the sur&ce of a rough sphere (radius «r) ; find equations for the movements of the rod and shelly there b&ng no fiiction Digitized byVjOOQlC 53 between them; and hence show that, if a particle fi is placed near the lowest point of a spherical shell (m) which performs small oscillations on a rough horizontal plane, the length of the simple equivalent pendulum is — ^ a. Solution by the Fbofossb. Let the horizontal and vertical lines through the centre O of the sphere be taken as axes of x and y respectively ; let (x, j/) be the coordinates of the centre C of the shell ; ^ » angle through which the shell has turned ; ^ « angle between radius and Ojf; wm COy; c » perpendicular from C on rod. Then « — (m4-/u)^8y + /ic^cos^S^. Substituting for x and y their values in terms of $, and selecting the co- efficients of 80 and 8^, we obtain, by equating these coefficients separately to zero, the equations (2«+M)(a+0S-^[008(*-<')g--(*-»)(g)'} + (m+ii)gcoae — 0, c(a+r){co8 (♦-») g + sin(^-e) gyj-(*»+o»)g +gc^4> - 0. In the particular case in which the radius becomes a partidei and the sphere a horizontal plane, these equations become ac rin 4. ^ - (i»+ e») ^ + cj sin ^ - 0, dP fir which give (by putting ^ — ^ir— iS, where jS is small) showing that the length of the simple pendulum » a. In the case in which the sphere reduces to a pkne of inclination » t, our equations become (2»+M) « ^ + /«: ^£?i^ + (»+m) J Bin .• - 0. «,rin(^+.-)^ -(*»+c»)^ + ^coe ♦ = 0. Digitized byVjOOQlC 54 222L (Proposed by W. S. Buskside, M.A.) — 1. Show how to deter- mine the locns of the feet of perpendiculars from a fixed point on the gene- rating lines of a system of confocal qaadrics. 2. Prove that this locns is described by foci of the plane sections passing through the fixed point. Solution hy the Pboposeb. 1. Since the foot of a perpendicalar on a generator is the foot of a per- pendicular on a tangent plane through that generator, the question ( 1) is reduced to first finding the locus of feet ot perpendiculars on tangent planes to one surface of the system, and then eliminating the arbitrary parameter which enters by means of the equation of the surface itself. 2. Consider the section of a surface of the system by a plane P, and let one of the foci of the section be at F ; this being so^ reciprocating origin at F, the section of the reciprocal surface by the plane P is a circle ; con- sequently the section of the asymptotic cone of this burface is a circle ; but the asymptotic cone is similar and similarly placed to the cone generated by lines passing through F perpendicular to the tangent planes through F to the given surface ; therefore one of the focal lines of this cone is the perpen- dicular to the plane P at F. Finally the focal lines of the tangent cone are the generators of the confocal hyperboloid of one sheet through the vertex. 2795. (Proposed by S. Tebat, B.A.)— Find the average square of the distance between the centres of the inscribed and circumscribed circles of a triangle inscribed in a given circle. I. Solution hy the Bev. J. Wolstenholice, M. A. If a be the radius of the given circle, * radius of an inscribed circle, square of distance between the centres is a^ — 2aa;, and the question is equivalent to finding the average value of x. Now, considering one point A of the triangle fixed, which is clearly allowable ; AO the tan- gent at A ; P, Q the other angular points of the triangle, zOAP«a, Z0AQ = 4>, (4>>a), ar = 4a8inie8ini(<^-e)cosi<^, /*' /'"^4fflni«8m^(^-0)co8^<?e<2^ .*. avei'age vahie of a; » a d0d(t> (See Book of Mathematical Problems, Ex. 1036); therefore average square of distance between the centres is Sa^ M — _ J. Digitized byVjOOQlC 65 II. Solution hy the Pbofosbb. Let D be the distance between their centres ; then =^ R2 (1-8 sin ^A sin JB sin ^C) «R2-4R2siniA {sin^ (A + 2B)-sin ^A}. Supposing the vertical angle to remain constant while B varies from to ir— A, we find the sum of the squares of the distances = 2 /*' f*'^ D2<2AdB « 3R2(ir«-8). o o The number of triangles — 2 /*' j dk d^ = ir^. o o Hence the average square of the distance » 3R^ / 1 j. 2797. (Proposed by the Rev. J. Wolstenholmb, M.A.)— In any com- plete quadrilateral, the radical axis of the three circles whose diameters are the diagonals will pass through the centres of perpendiculars of the four triangles formed by the four straight lines. Solution hy Matthew Collots, B.A. Let ABCD (Fig. 1) be the quadri- lateral; then the circles whose dia- meters are the three diagonals AC, BD, £F will have their centres at the middle points G, H, K of these dia- gonals ; the three altitudes of the tri- angle ABP, viz., AOA'. BOB', FOP', meet in one point O (its ortho-centre) ; therefore G^, perpendicular to AOA', is parallel to BC and bisects AA' in g\ similarly, HA perpendicular to BB', and Kk perpendicular to FF^, bisect BB' and FF' in h and *. Fig. 1. Now, the square of the tangent drawn from O to the circle (G) = 0G2- AG^ - Og^-Xf = OA . OA' ; similarly, the square of the tangent from O to the circle (H) = OH^-BH^ - Oh^-Bh^ - OB . OB', and the square of the tangent from O to the circle (K) = OK^-FK^ = 0;fc«-F*« -OF. OF'. But OA.OA':*OB.OB'«OF.OF', therefore the tangents from O to the three circles are equal to each other, consequently the ortho-centre O of the triangle ABF is a point on the radical axis ot these three circles. For like reasons, the ortho-centres of the other thr^ triangles formed by three Digitized by Google 56 other ddes of the quadrilateral ABCD (yu., the triangles ADE, BCE, CDF) He also upon the same radical axis. CoBOLLABT.— Ag the three circles (G), (H), (K) are proved coaxal, their centres, viz., the middle points of the three diagonals of the complete quadri- lateral ABCDEF, lie in one straight line. A rimple proof of this theorem may also he ohtiuned as follows : — Let ABCDEF (Fig. 2) he the com- plete quadrilateral; AC, BD, £F its three diagonals. Through B,C,D,E,F draw lines parallel to the sides of the angle A, forming the parallelograms ABGD, AEHF, &c Then (hy Euclid I. 43), parallelo- gram CK»AC»CK', nnd therefore parallelogram LE»L'E'; hence (hy Euclid I. 48, ex absurdo) the points C, G, fl lie in one straight line. Now, as the diagonals of a parallelogram hisect each other, the middle point of the diagonal BD is the same as the middle point of AG ; and, for a like reason, the middle point of the third diagonal EF is the same as the middle point of AH. But the middle points of AC, AG, AH lie plainly in one straight line parallel to CGli ; therefore the middle points of the three diagonals AC, BD, EF of the complete quadrilateral ABCDEF lie in one strught line. Note. — Several new and curious theorems can he easily ohtained from the foregoing theorem and its corollary hy means of the methods of reciprocation. and inversion. I ,1.0-^1 •A <^h I' / U -"// V r Fig. 2. H (Proposed hy A. W. Pauton, B.A.) — ^The equation connecting the distances (ri, r^ r^ of any point on a Cart^an oval from the foci is ifi-i) o*ri+ (7-a) i8*r2 + (a-jS) -^r^ = 0, where a,fi,y are the distances of the foci from the triple focus. Solution hy W. S. McCay, B.A. The Cartesian oval may he written Zri + wrj+wrs— 0, or Xfi + fira « d (1.2). Now to reduce the first of these to the second hy means of the identity An2+Br2«+Crj3=:D (3), (where A, B, C are the distances hetween the foci, and D =:: ABC) we get hy suhstituting r^ in (3) from (1) Ar^^-^ Bra« + C (^I^V- D. Digitized by GooqIc 57 That this may be identical with (2), we must have P . m» An»+CP .(6,6). ...(7). ABC B»?+Cj»* A5 B* The last eqnation becomes by means of (5) \^ I /i^ ^ — : — Now since (2) may be written (.v..-.V-^5V^'(n.-..,^)-o. it follows that if (a, jS, y) are the distances of the foci from the triple focus ^centre of circle xVi"-/*^ r^^ « ^4)' ^'«-m'^ * 0. Hence (7) becomes ? : i» « Aa* : B/B* ; so too m : i» = B/S* : Oy*; therefore llmln^ (fi-y)^ : (y-a)fi^ : (a-0)y^. 2670. (Proposed by C. T. Hudson, LL.D.)— An observer, seated in the aisle of a cathedral and looking westward, sees the horizontal lines above the arches converging to a point before him, and consequently cutting the horizon at a certain angle. On his facing northwards, he sees that the same lines are now parallel to the horizon. Required the curve that they will appear to lie in, as he gradually turns his head through 90^. SoluHon by tXe Pboposes. This problem is indeterminate ; for the plane on which the given lines are seen projected at any instant, will at that instant be referred to various distances according to the fancy of each observer. (i.) Let this plane be supposed to be vertical, and always at a fixed distance from the observer; then the ultimate intersections of its various positions will be a vertical cylinder of which the axis passes through the observer, and a plane through bis eye and the given horizontal lines will cut the cylinder in an ellipse, which therefore is the required curve. (ii.) Let that portion of the given horizontal line which is seen distinctly at any instant be supposed to be referred to its real distance from the ob- server at that instant. Let O be the observer, AB tlie given horizontal line, Oy and Ox lines perpendicular and parallel to AB in the plane drawn through AB and O, QPthe portion of AB which is seen distinctly when the observer is looking along OP, but which is seen projected into the position PR at right angles to UP. Let also AO^^, POa?=0 ; then the equation to PR is scoaB + ysinB ^ccoaecO; Digitized by Google 58 and the nltunfttemtenectioiis of all its poeitioiis obtained by varying lie in the cnnre a* =» 4c (c— y) ........* (1). The sensation of the enrvatnre of Afi is, according to this hypothesis, pro- duced by the attention being fixed on the extremity of the line BP, which always moves so aa to to«cfa the parabola (1), and at the same time remain at right angles to the visual ray. 2762. (Proposed by the Esitob.) — Find the envelope of an ellipse which has one vertex in the curve of a given parabola, and touches at its adfaoent vertex the extremity of a fixed double-ocdinate of the parabola. I. SoImHoh hy J. J. WxLKBB, M.A. Taking the ordinate of the parabola and the axis of the ellipse which passes through the vertex of contact as axes of y and x, the equation to the parabola will be (n being the fixed ordinate) ya-2ny ■¥px « (1), and that of the ellipse y' +-^!^ + -?PiL«0....(2), where y is the ordinate of the parabola passing through the first vertex of the ellipse, since the axes of the ellipse will plainly be ^ + » and Equation (2), when multiplied by (]/''—»')' and arranged by powers of y*, becomes (j/2 + 2par)y'2_2«yV+ nV+P'-^^-Zpna* « 0, the discriminant of which, with respect to y* as variable, is (y^ + 2px) («^2 +pia^^^n^x) — nV» or i^a?^ (v^ + 2px—4,n^. Hence the required envdope is the parabola y^ + 2px—4!h^=^ 0, which has as ordinate the fixed double ordinate, and a parameter double that of the given parabola. II. Solution ly the Ret. J. Wolst£NHOLME, M.A.; B. Bills,* and others. The equation of the parabola being y- = 4aa:, let the coordinates of the fixed vertex be <ia?, 2aa, and of the moving vertex aA-, 2a\ ; then the equa- tion of the ellipse is i^Zf^)!_ + (y-2^«)' ^ i or 4 (x- «X3)2 + (o + \y (y - 2aaf « 4a2 (a^- \^)^, Digitized by Google 59 or 4(ar-aa5)(ar + ai^-2aX*) + (a + A)*(y-2aa)«=: 0; and the enTelope is a2(y-2aa)< = {(y-2aa)«-8a(ar-ai^)}{<^(y-.2aa)2 + 4(a>-aV)], or (x-a«^»[(y-2ao)«-8a(ar+ao=)} « 0. The required eavelope is thtis seen to consist of the fixed doable ordinate, and a parabola similarly sitnated to the given one, of twice the dimensions, and having its vertex in the diameter through the given vertex. For the contact to be real, the vertices must be on opposite sides of the axis of the parabola. 2800« (Proposed by J. Wilson.) — Three circles passing through a point P form a circular triangle ABC, and each side of this circular triangle or its continuation is cut orthogonally at the point P' by a circle passing through P; prove that the three circles described about the triangles PAP, PBP", PCF" are co-axal. Solution hy Abcheb Staitlit. By inverting the figure with respect to the point P, it will really be found that the theorem is the inverse of the well-known one, that the per- pendiculars are concurrent which are drawn from the vertices of a triangle to the respectively opposite sides. 2688. (Proposed by J. Gbiffiths, M.A.)— A variable triangle circum- scribes an equilateral hyperbola, and is such that its wine-poiwi circle passes through the centre of this curve ; prove that the locus of the centre of its eircwnacribing circle is the hyperbola in question. Solution hy the Rev. J. Wolbtbnholhi, M.A. If an equilateral hyperbola be inscribed in a triangle, the centre of the hyperbola lies on the polar circle of the triangle : if. then, the centre lies on the nine-point circle, it must lie on the circumscribed circle, since the three drcles are coaxal. Let O be this centre, ABC the triangle, and let the asymptotes of the hyperbola meet the circumscribed circle in P, Q. Then, since OPQ, ABC are triangles in the same conic (circle), their sides will touch one conic ; but five of them touch the hyperbola, therefore also the sixth (PQ) touches the hyperbola: it will be bisected in the point of contact, which will therefore be the centre of the circumscribed circle of the triangle* Hence, if a rectangular hyperbola touch the sides of a triangle, and its centre lie on the nine-point circle of the triangle, the centre of the circumscribed circle lies on the hyperbola; which proves the theorem. Digitized byVjOOQlC 60 2654. (Proposed by W. S. Bubkbidb, M.A.) — Determine the polar reciprocal of the quartic curve ^ + «V+a^3+2iiy«(Lp+my-HftB) « 0. Wnte J. 1. I SoUaion by the Pboposkb. for X, y, iB, and the qnartic 18 transformed to X>+T«+Z2+2nrZf2«iZX + 2«XY = 0, and- the polar of a pomt ajSy becomea aTZ + jBZX + 7XT ^ 0. So the qnestion is rednced to finding when these two conies touch, which is easily done. 2726. (Proposed by the Rev. J. Wolstenholkb, M.A.) — If a conic touch the ndes of a triangle and pass through the centre of the circumscribed circle, the director circle of the conic will touch the drcumscribed circle of the triangle. Solution by W. S. MoCat, B.A. If the conic be (te)* + (wty)* + (n»)* — 0, it follows from the general equation of the director circle given at p. 338 of Salmon's Conies ^ that the radical axis of director circle and circumscribing circle of the triangle is icot A.a? + flicot B.y + ncot C.» i» 0; and if this touch the circumscribing circle {ayz + hxt + cxy — 0), we must have Pcoe? A + »i»cos» B +n«co8« C— 2m«ri^ cosB cos C - 2Pn^coB B-2m«i«cos2 C - 0, or (I cos A)* + («» cos B)* + (» cos C)* =■ 0. But this is the condition that centre of drcumscribmg drde (cos A, oqb B, 008 C) should lie on the conic. [Mr. Wolstenholmb's proof is as follows :— Let O be the centre of the circumscribed circle, and let a tangent to the conic at O meet the circle in B', C; then the tangents from B^ (^ to the conic will meet in a point (A') which will lie on the circle, since ABC, AfWQf are triangles circumscribing the same conic These tangents being at right angles. A' is a point on the director circle; and since WCf is bisected in O, OA' passes through the centre of the conic, that is, through the centre of the director drcle. The <drcles therefore touch at A'.] 2770. (Proposed by the Rev. J. Wolbtenholme, M.A.)— If a conic be drawn through S, S', the foci (both real or both impossible) of a given conic, and osculate the conic at a point P, the tangents at S, S' will intersect in the centre of curvature at P. Digitized by Google 61 SohOion by tie Fbofossb. The equation of any conic touching the g^ven conic ^ + IL b 1 at the point (a cos 9, ( sin 9) is if this osculate at the point $, L cosi^ 9 + M sin* 9 a> K ; if it also pass through the two fbd on the axis of x, L+N«0, A-5:!Vl + Loo8«9)-l-N; Also the pole of SS' is given hy the equations 0+Lc(rfa)f+°°''=°'*(L a o But, from the former equations^ 2(l+Lcoe»9)*+??L?i5^(L + M)y-0, M±?5r8in9+2a-N) « 0. a o o -L-N=. ** - ^«»°*^ and a3fflnS9+^oosS9 l + cu8S9* M + N 2A« ,^_^ a»-ft» , 1-N (a«-^8in<fl L + M 252co8«9 therefore ?-^^ ^ . - ^ . .^ therefore y =- iJZf: rin» 9 ; *9^ "^ 6 1 + LC0829 a2sin<9* therefore f - *i?!!;! . ^ sin' 9 . ^ cos^ 9 ; a a'sin'9 o a' or the tangents at S» S' intersect in the point (^ " cos* 9, ""^ mn* 9], the centre of curvature at P. It may be noticed that, in any conic passing through SS' and touching at P, the tangent and normal at P, form with PS» PS' a harmonic pendl ; therefore SS' and the normal at P are conjugates with repect to this conic, or the tangents at S, S' intersect on the normal at P. 2799. (Proposed by A. Mabtin.)— If n dice are thrown, what is the chimce of an odd number of aces turning up ? Soluiion hy the Rby. J. Wolstbnholkb, M.A. The chance is obviously . If the dice be regular tetrahedra, the chance is —> 2 2""*"* Digitized byVjOOQlC 62 2789. (Proposed by W. S. M*Cay, B.A.)— If two drdes pass through the foci of an ellipse and touch the same variable tangent to the curve, the angle at which they intersect is constant and equal to 2 tan~i -. Solution hy the Rev. J. Wolbtekholmb, M.A. Let SS' be foci of an ellipse, TPT a tangent meeting the tangents at BB', the ends of the minor axis in TV, and the major axis in U. Then ZUTS'« Z BTS = I TSS'; therefore triangles TITS', UST are similar, and UT : US' = US : UT, or US.US'-UT2; therefore the circle through STS' will touch the tangent at T, so the circle through ST'S' will touch the tangent at T^; and the angle at which these circles cut ift S, S' is the sum of the angles STS', ST'S', and therefore the difference of the angles TS'T', TST'. But Z TSr- i z BSB', and Z TS'r = J external angle BS'B'= w- |BSB'; therefore angle at which the circles cut « ir— BSB^= 2 tan-i£. The Brst part of this problem I have already set in the form, " If a circle pass through the foci of an ellipse, and common tangents be drawn to the ellipse and circle, their points of contact with the circle will lie on the tan- gents to the ellipse at the ends of the minor axis." 2780. (Proposed by Professor HiBST.) — The envelope of the chord com- mon to an ellipse and its circle of curvature is a curve of the fourth class which has three double tangents, one at infinity and the two others coinci- dent with the conjugate diameters equally inclined to the axis. Prove this, examine the curve, and consider the cases of the hyperbola and parabola. Solution ly J. J. Waleeb, M.A. The equation to the ellipse being b^x^-^cfiy^ = a%'^, if the coordinates of any point on it be a cos 0, b sin 9, the reciprocals of the intercepts made on the axes by the chord common to the ellipse and the circle of curvature at this point will be a = -5?L — , fi « -Zfl!l__, and eliminating between ^ acos2e hem 29 ^ these equations, a^a^ + b^ff^ = (a^a^—b^fi'^^ is the envelope of the chord, — a curve of the fourth class, having as double tangents the lines joining the points aa±bfi = with the centre, — i, e,, the equal conjugate diameters; also the line a=0, jB=0. For the hyperbola, the sign of 6^ is to be changed, and the first pair of double tangents become imaginary. In the case of the parabola y^ =■ 4mxy the tangential equation to the envelope of the chord common to it and its circle of curvature at any point Digitized byVjOOQlC 63 is nmilarly found to be « » Zmfi^, a parabola having the same vertex with the given parabola, and as its axis the production of axis of the other. Its parameter is three times that of the given parabola. The equations to the above envelopes in Cartesian coordinates are (AV + aV-4a262)3+27(Z^2j^-aV) = and f/^ + l2mx^0 respectively, the equations to the ellipse and parabola being b^a^ + a^y^ = a%'^ and y2 «■ 4mx respectively. From the former it -appears that the points of contact of the equi-conjugate diameters lie on a similar and coaxal ellipse, and that there are cusps at these points. [The equation of the envelope has been given by Mr. Wolstenholme, in his Solution of Question 2579 (2), (Reprint, Vol. X. p. 91) in the neat form (M)'^(M)*=^' which has cusps at the points 5- s= ^ » 2, and of course two at infinity. The form of the equation to the envelope given by Mr. Waleeb in Question 2834 may be readily deduced from Mr. Wolstenholme's.] To express the distance between the centres of the ci/rcumscrihed and inscribed circles in terms of the radii of those circles. By J. Walmsley, B.A. Let O', O be centres of circumscribed and inscribed circles respectively of the triangle ABC. Join O'O, OB, OC. Produce BO to D, and join DO'. Di-aw O'P pei*pendicular to BD, and therefore bisecting it. Tlien B0 = . and therefore Hence therefore sin-^B DCO = DCA + ACO = DBA + AGO - OBC + OCB = DOC 5 asin^B ^ 2B$ia^B. DO = DC=^ . ^ Bin A DP2-0t»* ^ DO . OB - 2Rr. 0'D»-0'02 0'0« « R2-2Rr. In an analogous mannei^ if Oi be eentre and ti radius of escribed circle touching BO, we may show that 0'0i2 = R2 + 2Rri. 2776. (Proposed by W. K. Clitfoed, B.A.) — Through A, the double point of a circular cubic, draw AB perpendicular to the asymptote ; if chords be drawn to the curve subtending a right angle at the double point, show that there is a fixed point in A B at which also they subtend a right angle. Digitized by Google 64 Solmihn hf the RxT. J. Wolbtevhoucx, M.A. The equation may be taken y^^a? ^^^i the origin being o— a? the double point. If (I'lyi), (jts^i) be the ends of two chords AP, AQ at right angles to each other, XiX^-^-yi^^ » 0; and h—xi 0— a?, therefore (a?i + a) {x^ + a) « (6 — tj) (6 — a?j), therefore «i + a^ = b — a. Bat the equation of the circle on PQ is (a?-ar,)(ar-a!^ + Of-yO(y-ya)=0, or a»+i^-a?(jri+«i)-y(yi+ya) - 0; therefore, when y =» 0, « » or a? «> a?i + x, » 6— a, giving a fixed point on AB at which PQ subtends a right angle. 2544. (Proposed by Professor Ball.)— There are two values oid for which a, ^, 7, the roots of the cubic a** + Zlv^ + 3cj: + d — fulfil the linear relation Aa + BjB + C7 •- 0. Show how to find the quadratic equation of whichlthese values of d are the roots. I. Solution htf the Bby. J. Wolstxkholmb, M.A. The relations a + i3 + 7- -_, Aa+Bi3 + C7 = 0, fiy-^ya+afi ^^ a a Icad^at once to a quadratic equation in 7 ; whence 86 A«+B2-C(A + B) _o ^*'*"^«""TA2 + B* + C»-BC-CA-AB=^ 8ac(A-B)3+ 9ft-AB-_ '^»'^»'*aHA»+...-BC-...;-"- But -(c?i + <y= +a(7i8+73»)+3ft(7i2+72S) + 8c(7i + 72), and didi = 7172 (071' + Sbyi + 3c) (0732 + 3672 + 3c), which can be immediately expressed in terms of m and n. II. Solution hy J. J. Waleeb, M.A. By the terms of the question, BjB + C7 = -Aa (1), afi-^ay ^~-aa-Zh (2), a^7 + a (i8 + 7) a « 3c (3). From (1) and (2), a(B-C)i8 - a(C-A)a + 36C (4), a(C-B)7«a(U~A)a + 36B (5); Digitized byLnOOQlC 65 -a5(B-C)»jS7 - o»(A-B)(A-C)a» + 3aA{2BC-A(B + C)}a + 95»BC (6), Eliminating /3 and 7 from (2), (3), (6), there results a»[(A-B)(A-C)-(B-C)«}a»+3a5{B« + C«-A(B + C)}a + 3[ac(B-C)« + 362BC]«0 (7); also aa» + 3*a' + 3ca + (J « (8). The resultant of (7) and (8) is a'^d^ + 9a' { (3^»'2- 2aV) (a'J - aJ') - a'U {a'c- ac') } d + 81c''(a'J-.aft')H27c'(a'c-.aO{a'(a'c-aO-3J'(a'6-a4';} -0... (9), if we write a'= aa{(A-B)^A-C) + (B-C)»}, &'-=a6{B2+C2-A(BfC)}, c'=ac(B-C) + 362BC, whence o'6-ay=a34(A2+BC), a'c -oc'- a {ac (A- B) (A- C)-362BC}. Making these substitutions in (9), it becomes divisible by a\ and is the re- quired quadratic for tf. The coefficients are not capable of any important rodnction, as far as I have examined them. 1587. (Proposed by the Eev. T. P. Kiekman, F.R.S.)— Apollo and the Muses accepted the challenge of Jove, to vary the arrangement of themselves on their fixed and burnished conches at his evening banquets, till every three of them should have occupied, once and once only, every three of the couches, in every and any order. In how many days, and how many ways, did they accomplish the feat, keeping one arrangement of themselves through all the solutions P Required two or more ot these solutions, clearly indif* Gated, so as to save space, by cyclical operations. Soluiiafi hy the Pbopobbb. The solution is the positive half of the group of 10. 9 . 8 . 2 made with 10 elements, which was first given by me in the Memoirs of the Literary and PhilosopMcal Society of Manchester^ 1861, and whose title I gave thus in their Proceedings five or six years ago : — 10.9.8.2 =» 1 + 180q2 + 14458 + 452418 + 80381 + 2704,1, + 802,14 + 18081, + 144io + 90^ +240ggi + 363, : Q - 2520. The positive half is the first six terms on the right, comprising, besides unity, the natural order, 180 substitutions having each a circle of 8 and a circle of 2, 144 substitutions having each two circles of 5, 45 having each four TOL. XI. H Digitized byVjOOQlC 66 circles of 2, and 2 elements undisturbed, 80 having three circles of 3 and one undisturbed element, &c. This group of 10 . 9 . 8 . 2, ns well as every group so given by its title, can be easily constructed from its title ; but it is impossible to state the rules here. It is enough to give one of the 2520 equivalent groups of 10 . 9 . 8, thus, 1234567890 1234567890 1234567890 +0324897561 3215648970 1473692580 +3017594862 3126459780 1327984650 +2107864593 4567891230 1742853960 +6280715394 5648972310 1685739240 +3814069275 6459783120 1956278430 + 8243907156 7891234560 1869425370 +8639520147 8972315640 1598346720 + 4261539078 9783126450 + 2165437809 under the form (A);G;H where G;H is a group of 9.8, the product of two groups G and H. (A) is no group. The 10 . 9 . 8 cannot be written as the product of so few as three groups. We have the g^oup by adding to G;H its nine derivatives by the substitutions of (A). If these be 1, Ai« A2,. . . . A9; ly Sfh Sf2' " * Sfs those of G ; and 1^ ^. . . . A7 those of H ; the group (A);G;H = G;H + Ai;G;H + Aj;G;H+ .. .. + A9;G5H, will be found to stand the test, that every product of its substitutions, if 1 =■ Aq = ^0 = ^o» is of the form A ; ^ ; h^. Neither the algebraic nor the tactical demonstration of this can find space here. The group proves itself. These 10 . 9 . 8 arrangements are one of the 2520 solutions of the problem which Apollo's party dined out day b}"^ day for nearly 5000 years, at some expense of Jove's nectar. There are three, and only three, solutions which have in common the substitutions of G, (A);G;H, (A');G;H', and (A");G;H". No two solutions have more than eight arrangements in common besides unity. The above group (A);G;H is necessary as to the number, and sufficient as to the form, of its arrangements : necessary, because Apollo, Clio, and Urania cannot seat themselves in every way in fewer than 10 . 9 . 8 arrangements ; and sufficient, because if in e and *, any pair of arrangements E, L, and M were seated alike, the group would contain the substitution — , which has KLM undisturbed; but the positive title has no substitution showing three elements undisturbed. From (A);G;H once written out can be formed all other solutions, by the operation 6;(A);G;U;0'^, where $ is any substitution not before used. Thus, e ^e-' may be 2134567890 for a second solution. Note. — The unfashionable reader who wishes to learn how to find all possible groups made with n elements, to determine their titles and the number of their equivalents, and from the titles to construct the groups by a direct tactical method, without troubling himself with congruences, has only to fish up, if he can, from a basket in Manchester^ a Memoir of mine which has lain there for some years. Our learned societies agree with me in the opinion that it is absurd to waste English paper on sudi upstart subjects as Groups and Polyedra, which are fit for nothing but make-believe prize- questions of the French Academy. They have done quite enough in phicing it beyond dispute in their Proceedings that I have thoroughly discussed both these vast theories, as difficult as they are unvalued. Digitized byVjOOQlC 67 2847. (Propoied by M. W. Ceoftok, F.R.S.)— Prove that c*D\ «-*«• - (l + 4A*)-*€"i"^, where D^~ dx SolnOian by J. J. Walkbb, MJl. Expanding the symbolical operator, ^D..,-*^^n.*I>'.*^. -te" But by the known formula for the saccesdve differential ooeffidents of c with respect to x, €*^ A D«. €-*** = « ^ A* + A*2 (2x)\ (1.2.3)» (1.2)2.3 ^ ' ^ ^A»*»(2^)4+ *!^(2*)«, 1.2.3 " ' 1.2.3' It will be found on examination that the rth term (or line) of the ^h column in the sum of the above developments is equal to ^ ^ 1.2...(j,-l) ^ (2r + 2p-4).(2r-f-2p-5)...(2p-l).(4Alr)'"^ 1.2...(r-.l).r...(r+p-2).p.Cp + l)...(r-l).(2)^-"' when r is greater than p ; but when r is equal to or less than p, the denomi- nator of the last factor is 1 . 2 ... (r— 1) . p . (ji + 1) ... (p + r - 2) . 2^"*. In both > this factor is equal to (^^" ^) (2p + D - - (2p + 2r- 6) (4^;t,)r-i ^^^.^ 1.2...(r-l).2*-* multiplied by (— l)*"'^ is the rth term in the development of (1 + 4Ak) ' . From this it results that f*^. €*»". €"**■= (l + 4Wr)-*c^****j and multiplying both sides of this by c" , there results for c . c" the value given in the [This theorem, which the Proposer has found to possess important appli- cations in the Theory of Errors of Observation, seems deducible in the most direct, though not the most elementary, manner, by putting ^'^^ for ^(x) in Poisson'b famous transformation {TraiU de Micanique, Tom. II., p. 356)^ which gives ^{r)^l-f €-V(* + 2A*«)rf«.] Digitized byVjOOQlC 68 (Proposed by W. S. McCat, B.A,)— If a conic pasg throagb the foar points of contact of tangents to a cabic from a point (A) on the curve, and throa$?h two other points (B, C) on the cubic ; then A is the pole of BC with regard to the conic Solution by F. D. Thokbok, M.A. The equation to any cubic may be written S=jTV+*yU = (1), where U, V are conies intersecting in four points A, B, C, D on the curve, and a and y are straight lines intersecting on the curve, in the point O suppose. The equation to the polar conic of (1) with reference to the point (a?', y', z") is / dS J dS / dS ^ dx dy dz which may be written a?^ + VU + ^Pv+*^P|* = ^ (2)» where P^ denotes the polar of (j:',y, z') with respect to U. But if (ar', y', «') be on the line OP which touches U, a/ss 0, and P becomes a?-^. Hence the equation (2) reduces to Ar/U + ar(p^+^g,)-0. Therefore, if U is itself the polar conic of the point P, P +>ty^=0 identically. dor Therefore P^ coincides with y — 0, ».e., with EP, or EP is the polar of P with respect to V. • 2606. (Proposed by C. W. Meeeipield, F.R.S.)— The developable cir- cumscribing two surfaces of the second degree touches either of them along a curve, which is its intersection with another surface of the second degree. I. Solution by the Rev. R. TowKBEin), F.R.S. If U and V be the two quadrics, U' the polar reciprocal of U with respect to V, and V the polar reciprocal of V with respect to U ; then, for every plane tangent to both U and V, the point of contact with U being a point on V, and the point of contact with V being a point on U', therefore, &c» (See Salmon's Geometrif of Three Dimensions, 2nd Ed., Art. 207.) If aa^ + JiB« +073 + d82 ^ q and aV + b'e? + c V + d'h^ « be the equa* tions of U and V referred to their common self-reciprocal tetrahedron, then are Digitized byLnOOQlC 69 f!,.+ ^V+^7»+?«'-0 and li:«'+^/,.+ ''-y.+ |».-0 abed a V c of those of U' and V referred to the same tetrahedron ; which, as is otherwise evident geometrically^ is consequently self-reciprocal with respect to the whole four surfaces at once. (See same. Art. 206.) II. Solution hy the Fboposeb. Referring the two surfaces to the centre of one of them, and to axes parallel to conjugate diameters in hoth, their equations may he written thus; — U - aiar,«+iB,yi» + 7i2i'' + «i = 0, The tangent plane of U is aia?iaf + /Biyiy + yiZiZ + Jj = 0. The tangent plane of V is 03(a?2-0(a?-0 + /32(y2-«»)(y-»n)+'ya(«2-«)(2^-«) + 8* - oa 2* - /32 m' — 72 »' - 0, or OsCar— Oay + ZSjCyj— iii)y + 'y,(«j— n)« + J2-a8te2-iB2«yj-7i2fiM!j = 0. And, if these are to he identical, «i^i ^ ^lyi ^ 71^1 ^ ?} ojCaj^-O 0iiy2-fn) 72(«^J-») «2-a2(^2-/32»»y2-72W Suhfliituting those values ofxj^yiZi in U, we get «2'gi(^2-0» ,. /32'8i(y2-«>)' ^ 72^81 (g2-«y «i i3i 7i + (82-«2^2— /32«y2-72»»«2)' = 0. Hence the developable of two quadrics touches each along a curve through which another quadric can be drawn. Conversely, the envelope of the tangent planes to a quadric, along its in- tersection with another quadric, is the developable circumscribing the first and some other quadric. 2804. (Proposed by S. Tebat.) — A strdght pole stands vertically on a slope inclined to the south. If it be broken at random by the wind blowing in a given direction, so that the upper end of the pole rests upon the slope, determine the probable area of the triangle thus formed ; and deduce the result for a horizontal plane. Solution hy the Pboposeb. Let a be the length of the pole, x the length of the portion standing, a the inclination of the slope to the vertical, j9 the azimuth of the wind measured on the slope, and ^ the angle between the standing piece and the base of the triangle. Then cos ^ » cos a cos i3, and A = ia?8in^{afcos^ + (a2— 2aa:+«'c0B'^)*}. Digitized byVjOOQlC 70 The gmtflst valae of « if f — , which is also a menBore of the number 1 -|- BID^ of trianglee. Hence we find y ^"* 5 (8(l+8in4»)» 8co8?^ 2oo(i*^ 2coe»^^ "^i* an^ therefore the average area of the triangle is ia>8in * a + «n A) l_£?li 1 + ^ ''"'^ lojr cot 4*1. • '^^ ^M3a + 8in^)» 3oo8>4»^2co8<^ 2 cot* ^^ ^^3 When a«iir, we have ^»^; and expanding logcot^ ■• ^log±^l5?^> 1 — COB^ we find the average area « ^aK 2618. (Proposed by the Bev. K. M. Fibsbbs, M.A.)— A pack of n dif- ferent cards is hdd, race downwards, on a table. A person names a certain card, that and all the cards above it are shown to him and removed ; he names another, and the process is repeated. Prove that the chance of hia naming the top card during the operation is liV H* E" Solution hjf Pbofbssob Whitwobth. Let P^ represent the chance of the top card being named in the series of operations on n cards. At the first operation any of the n cards ia equally likely to be named. If the top one U namedy what is required is done, and the ii posteriori chance of success is unity. j[jf the second be namedy there are «— 2 cards left, and the d posteriori chsince of success is P„.2* J^the third he named, then the d posteriori chance of success is Pn.a, end soon. If the last card be named, the chance is zero ; therefore or i»P„ = 1 + Pi+Pj+ .... +P„.3+ P,_2- Similarly (writing «— 1 for n) (»-l)Pn-i « 1 + Pi + Pa+ .... +P„.85 therefore, by subtraction, nP^— (»— 1) P^_i - ^ii-2» »(Pn-Pn.l)=-(Pn-l-Pn-2)- Digitized byVjOOQlC 71 MaUaply both ndes by (—1)" |»-1 . then (-i)"lji(P»- P.-i) - (-1)""' t:l(P«-i- P»-8)- Now, writing n^l, »— 2, ii^3, &c. saooessively for n, we obtain, ex equdU, (-!)• l«.(P,-P,_i) - tl(P,-PO. Bnt it is obvions that Pi*l and Ps»i, therefore (-1)" WP„- P,_l) - -1. or Pn-P.-!' .(-1)-' Writing 2, 8, 4^ &c. saccesmvely for », we obtain P,-P.=._i-, P,-P,= + |., P,-P,»_l 4c P-P .- (-1)"-' Therefore, by addition, 11 11 11 t With the notation of Questions 2637, 2648, this chance (of naming a top card) is l^e^^, and the chance of not naming a top card is e'^ • 2813. (Proposed by J. J. Waleeb, M.A.)— The equation of a conic referred to an axis and tangent at vertex being ax^ + bi/^ + 2dje * 0, if the conic be turned about the vertex in its own plane through a right angle, the locus of the intersection of any tangent in the original position with the same line in the new position of the conic is the (bicircular) quartic and the corresponding locus for the normal is the sextic What does this latter equation become in the case of the panibohi ? I. Quaiermon Solution by W. H. Latxbtt, B.A. 1. Taking for simplicity the equation to the conic in the form 6V + ay-2a^«, the vector equations to the tangents are p «■ a + ocosT+i3 8inT + jr(— asinr + iSoosT), p » ?jS(1+cost) aunr+y (^i3sinr+ -a cost]; D a \o a J Digitized byVjOOQlC 72 and that these may coincide, we mnst have a? (a^ sin' t + ^ cos' r) — a (a sin t + 6 cos t) + (a'— 6^) sin r cos r + a& ; whence substituting, and putting o'sin'T + i'cos^T «= P, we have for the required locus f».P Bs [fta(6cosT— a8inT) + a/B(6cosT + a8inT)J (1 + cost). To transform this to a Cartesian equation, we have ar.P , . J v.P , ■ ttdcosT— asmr, and iJ . = ftcosT + asinT ; ab (1 + cos t) ab (1 + cos t) therefore ^^y. ^^-'^il^co-rf (,j^ also (pB-yY P3 - 4a2&2 (i + cos t)« o' sinS r or (a?-i/)«P =2a«8in«T(a?» + i/2) (2), and (* + y) P — 2a6 (1 + cost) 6 cost; from(l), {«»+5^-2a(ar+y)}P- 2a2is[(l + cosT)3-2oo8T(l + co8T)} = 2a2ft2 8in«r; from (2). {*3+y«-2a(a:+y)} {x^ + y^ = j^^oWy^+j^ ^ 6»(a?-y)^. or J {a(a:8+y3) + 2rf(a?+y)} (x^+t/S) = d2(ar-y)2. 2. The vector equations to the normals are p :^ a + a cos T + jS sin T +x(b'^a cost + a^fi sin t), f»'« -jS+ ?/BcosT asiuT +y(ha$ cost— aJasinr) ; ho a and that these may coincide, we have — aAa?(a*8in'T + 62co82T) = afdcosr— aMUT] +a5 + (6*— a')cosT sinT; whence substituting, and putting Q?ein^r + b^coB^T^V and a2(i + cosT)-62cosT = Q, we get for the required locus ahFp — sinT Q[5a(asinT + 6cosT)— /Ba(ico8T— asinT)^. To transform this to a Cartesian equation, we have a?P — sin tQ (a sin t + 6 cos t), yP « sin tQ (a sin t— 6 cos t) j whence we easily find that ^ 4a2&a8ln2r<y P* 46in-*rQg(ag-yyco68T , ,_ ^^ * P« aS62l6sin«TQ^cos«T pa «« V» r» or b{a{x+ify+b(x^yy] {a(x^+i/^ + dix+y)y^ ia--by€P(xi~ff^. 3. In the parabola our equations become Pa? = <Q[a<-5} and Py = <Q[a« + j}, where P and Q now equal respectively aH^ + 6^ and ^ aH^ + 6*. Digitized byVjOOQlC 73 Let r^ a 5 ^ at ^alf^ (where I is the latuB rectum); then we have easily a?R^=:T(2T-0S and yR/ = t (2t + Z) 8, where K^4fr^ + P, and S = 2t3 + ^; therefore Z ^±1!? =. §, and -2T-:/i±i!; a?+y 2 ar— y therefore 4 ^1±^' =. i (^-hy)H2(^~y)2 II. Solution hy the Proposes ; R. Tuoeeb, M. A. ; and others, 1. The tangent heing {aaf + (Q a? + hy'y -{■da/=0, the equation to the same line, when the conic is turned through a right angle, will be iy'a? — (aa/ + d)y-da/^ 0. Adding these equations, we get (aO'+d) (x—y) = — 6/ (a? +y) ; and eliminating y by means of the equation to the conic, (aa?' + d)2(*-y)2 + 6(aF'3 + 2(^y)(*+y)2 = (1). Again, eliminating y' between the two equations above, we have (aay' + rf)(a;3+ya) + ^(a.+y)„0 (2). The elimination of a/ between (1) and (2) gives the first locus, which in the case of the parabola evidently reduces to a (circular) cubic 2. The equations to the normal in the original and new position of the conic are, respectively, iy'a?-(aa?'+«?).y+y {(a— 6)a^ + rfJ = 0, and (flw?' + d)a? + ^y+y{(a— ft)a?'+cfj - 0. Pursuing steps similar to those in the case of the tangent, the required locus 18 easily obtained. In the case of the parabola, developing, reducing, dividing by a, and finally making a^O, there results 2(jP-y)2{ft(aj'+y3) + d(»+y)} + (?(a?+5f)8-0. If the given conic be a circle; the locus evidently becomes a circle also. 2839. (Proposed by J. J. Waleeb, M.A.) — In a triangle the bisector of the base is equal to the less nde and also to one-half of the greater side; de- termine the three angles. Solution hy the Pbofobbb ; R. Tuokeb, M.A. ; and others. Let D be the middle point of the base BC, A the vertex of the triansle, and AC»AD»4AB. Then AC must bisect the angle between AD and BA produced (Euc. VI. A), and Z CAD « 180°- I BAG, TOL, zi. Z Digitized byVjOOQlC 74 therefore I BAC - 2C, therefore 3C » 180^ -B, whence Bin 3C ■> Ssin C — drin' C « nn B. Again, rin C : sin B « AB : AC » 2 : 1, or sin C « 2dn B, whence 6 sin C — 8 ain* C » sin C. IMsregarding the irrelevant solution CbO or 180^, we have sin' C « {^. But co8BACi«l-2sin3C«-^, and sin' B « i sin' C - ^3. 2818. (Proposed hy the Rev. J. Wolstxitholice, M.A.)— Given a circle S and a straight line a not meeting S in real points; O, O' are the two point-circles to which, and S, a is the radical axis ; two conies are drawn osculating S in the same point P, and having one focus at O, O^ respectively; prove that the oorrespon(ting directrices coincide. Solution hy B. TucKSB, M.A. Join OP, OT, and draw OK, O'K, perpendiculars to them ;. then it is plain that these lines intersect in K on the circle through O'OP, cutting S orthogonally; and K is the intersection of the directrices. From the centre S draw SL, SL' perpendiculars on OP, O'P, and liM, L'M' perpendiculars on SP; then (Salmon's Conies, Art. 243) OM, O'M' are the directions of the axes of the conies, and therefore SM « SP sin» OPS. SM' - SP sin^ O'PS ; SM sin' OPS OPa SO. SM' ^ sin«0'PS *" O'pa " SO'' that is, OM is parallel to O'M', and the ^ectrioes (the perpendiculars from K on the axes) ooindde. 2781* (Proposed by T. Cottbbill, M.A.)— Find the conic envelope of the radical axis of two circles, one of which is fixed, whilst the other passes through a fixed point. If the radii of the drcles are equal, find the foci and axes of the envelope. Solution by the Rev. J. Wolstenholmx, M.A. Let the fixed point be origin, a^+^'— 2ax— 2fiy a the equation of the moving circle, and (j?— o)*+y' « r^ that of the fixed circle; then, if r' bo the given radius of the moving circle, the radical axis is A4? + fiy "OX + i (a^- f^ « 0, with the conation x' + /*' - 1^1 Digitized by Google 75 and for the envelope or the envelope is a conic having^ its focus at the fixed point, its directrix bisecting the tangents drawn from the fixed point to the fixed circle, and eccentricity the ratio of the distance between the fixed point and the centre of the fixed circle to the radius of the moving circle. Tf the radii be equal, the envelope is the reciprocal polar of the fixed circle with respect to a circle whose centre is the fixed point and radius {i(a-— r^) j *. 2724. (Proposed by S. Txbat, B.A.)_Three smooth rings, P, Q, B, are thread on an endless string of given length, and constrained to move on three straight rods, OP, OQ, OR. To investigate the motion corresponding to a slight arbitrary disturbance of the system. Solution by the Pbofoseb. Let OP=fa?, OQ-y, OR = «, PQ = «, OR = V, RP « 10, 81, 83, $8 *he inclinations of OP, OQ, OR to the vertical; and T the tension of the string ; POQ-a, 0PQ«4», 0PR=:4»', QOR-iB, OQR = x. OQP = x'. ^r R0P = 7, ORP^tf. ORQ-f. / Then, for the motions of P, Q, R, we have P (^ - ^ cos 5i^ + T(co8 ^ + cos ^') - . Q (^ - /^ cos 82 ) + T (cos X + cos x') « . R (^ _ ^cos 8a) + T(cos 4^ + cosf ) « . And eliminating T, P(co8,»-+co6f) (g _ jcos 8,)-B(coB4. + coi*') (^ -i? «»•».)- (4). QCoM^ + cos*-) (gf -jrco. 8,)-E(co.x+co6x') (^-i'ooe %) - (6). These are the general equations of motion, but their complete solution cannot be effected except for small oscillations. The geometrical relations give tfcos^ -fycosaa or, o cos x +2^006/8 "«y, looos^' +d?00B7 »«, foco8^'+«coe7«af, woosx'-f-^oosa ">y, « cos t{/ f y cos jS — iE. Digitized by Google 76 Let a^b^c be the Tfilaes of w, ff, z, when there is equilibrium ; a', 5', & the ▼alnes of «, v, w; and A, K\ /i, /Jtf, y, ^ thoie of cos ^, cos 4»', cos x> oos x'» cos ^, cos 4^. Assume dr — a + dr', y — ft+y*, « = c+2' (6), i« = a' + ii', ««6' + «', w = c' + fi/ (7), where j;', y', tf^ u*, v\ u/ are small quantities. Now «3=:a^+^-2a?ycoso, B3«y« + a8— 2y«cosi8, io»««? + aj*— 2jkfcos7; or, neglecting small quantities of orders higher than the first, aV= (a^bcoBa)a/+ (5-acosa)y, JV« (6-0008^8)^+ (c-^cos^jz', c'lc/— (c— 00087)/+ (a— c cos 7)0?'. But a~ft cos a » a', &c., therefore vf^Kx'-^fiy, «/-/ty + //, icZ-yz'+AV (8). Hence, from (7), « = c^ + KaZ+iiyf^v « 6'+/i^ + k'z', 10 « c' + xV + f*'. Since «'+i/+it/«0, equations (8) give therefore (.+xog + (M^/)f + (v^.^)^ - 0. Also,from(6), ^«^. ^-^, ^==f?!. Substituting these in (4) and (5), neglecting small quantities of orders lugher than the first, and putting A = ^(..04(V/-co.,)_(^4)a.,,')^'. D - {^(x+V)+^(XK-coBy)j^-i; (X+A')-J-(^V-coe«). B' - {^ Oi + MO +|- 0»v-«»'8)} ii±A'_^0' + ^')4 ^'^""^")' D ={f 0.-M')4&.V-cos.)}^'_(1.4)(l.^a we find, after reduction, v + v' ' dt* v + Z * c?^ ==5rCPAco8 8i + EBcosJs) +^(PCcos8i+ RDcos^), Digitized byVjOOQlC 77 = ^(QA'coB 83 + RB'cos 8j) + ^(QC'cos 8j + RIV C08 J,). Ag^n, putting \ - QR(A + \0* + RPCu + fi7 + PQ(i^ + i/)9, IC = {Q(F + |/)2 + R(fl + /)2](PAC08jj + RBC08^ -R(A. + A')(/i + /)(QA'co8 83+RB'cos5j), Z - {R0i + A*T + Q(i' + i'')2}(PCcos8i + RDcos«,) -R(\ + x0^f* + fi')(QCco8Ja + Rl>'co8«a), *'- {R(X + V)a + P(i' + i/)2}(QA'co882+RB'co8 8s) -R(x + V)G^ + /)(PAcos8i + RBcos5a), r= {P(i^ + k')» + R(A. + V)'}(QC cos Jj + Riy 0088a) -R(A. + A')0* + /*')(PCoos8i + RDcob58)» wefinaUyget *(v + v)^-^(Ara' + Zy} « (9), »(''+»o^-j(*'*'+«y)-o (10). Hence, for small oscillations, the motion of the system is given by the simultaneoos equations (9), (10), and will consist of two independent vi- brations, of which the periods are where jf, rf' are the roots of the equation Vi? + (*— 2') 1 — '• 2642. (Proposed by C. W. MBfiBiFiBLD, P.R.S.)— If three surfaces of the second degree meet in four points, at each of which their three curves of intersection have a common tangent, these four points lie in one plane. Solution "by the Pbofobeb. Write the equation of the quadric in its general form P = ax^+hy^-¥c^-¥d^+2lyz-¥2mzx-¥2nxy + 2px-\'2qy-¥Zrtwm0t and use suffixes to distinguish the three sur&ces Fi, Fj, Fs* Taking our origin at one of the points, and making the axes of ao and y pass through two others (2A, 0, 0) and (0, 2Jc, 0), we get {2»0 and j9a ^afi, q^^lk, where h and Ic are independent of the suffix. Take, moreover, the common tangent at the origin for the axis of x, and we have r^ r^ fs eacU =0. Digitized byVjOOQlC I 78 The tangent planes at the origin may therefore be written at aihx + hiky ^0, o-Ao? + bjity ■■ 0, a^ + bJhy^O; and as these are to have a common intersection, we get «8 " ^ ' Treating in a similar manner the points (2h, 0, 0) and (0* 2k j 0)y for which the tangent planes are (or— 2/i) ah -{-y {2nh— hk) + z, 2mh »> and a (2nk- ah) + {y-2k)hk +z. 2lk - 0, we get the tother relations fli-g« ^ wi-wg ^ 2(ni--iia)A-(&i-b2)lr Os f»8 2njA— 5j* ii — Jji l\—h _ 2(»i— > ig)^— (fli — flg)A ^3 " /s " 2jij*~a,A ' and since flZlf? - hzh, these are all equal, and so is also ^IZ!!!. We thus get the following reductions in the general equation :— d = 0, r - 0, £» . £2 = ?? « -A, £1 « i» - i» « -*, ai og aa ai Oj Og and we may therefore write the three quadrics as Pi « aypi^ + huf^ + ci«2 -|- 2iii/« + 2mi«a? + 2n^xy-~ 2aihx^2ibiJcy — 0, Fj - aja?3+ p, = f!ii:f?x2+ 5i-:^y«+c3«2+ 2^1:1^ yz+ -0, 99 9 the only outlying term being cg. But Fa- !izZ? « gives ^fiZfl+^aS « q, or there is a qnadric through the intersection which degenerates into the double plane ««0. This proves the theorem, which is easily reciprocated. The investigation, moreover, shows that, if there be three such points, there will also be a fourth. 2771. (From WoLSTENHOLias*8 Book of Mathematieal Problems.)^ Prove that (1) if a straight line of length a be divided at random in two points, the mean value of the sum of the squares on the three parts is ^a'; (2) if the line be divided at random into two parts, and the longer part again divided at random into two parts, the mean value of the sum of we squares Digitized byVjOOQlC 79 on the three parts is ff a' ; and (3) if it is an even chance that n times the sum of the squares on the parts in (1) is less than the square on the whole I. Solution hy Stbfhek Watsoit. (1.) Let jf, y, and a—x—y he the parts. Then the total of cases is a«, and doabling because of the interchange of x and y, the average is o (2.) Here the number of cases is and the average is O (3.) In this case, let a?, i(a— «)— y» \{a^x) be the parts; then aj2 + i(a-ar)2 + 2y«<£! (1); therefore y< +i(TOa2-3«2)*, 2 2 where m wm ~ and % s x—^a. Hence the number of positions y can n 8 take for each value of x is (ma'-3c^^» and therefore the chance p of (1) being fulfilled is, since dx^dz^ therefore » -•- — ^—r^ (whenp=i)^ ■^^' 2ir + 3/>V'3' "^ '^ ' 4ir + 3V'3 II. Solution hy the Bey. J. Wolbtestholme, M.A. (1.) The answer to this pert; of the question is /** /*"{af»+(y-«)8 + (a-y)2} dxdy-^ T C^'dxdy • c - 2a» /'^{(l + arS)(l-a?)-(l + fl?)«(l-*3) + |a-«»)}cfaf -*'»^/'\2-6« + 9««-5«») - 20^(2-8 + 8-4) - ia«. Digitized byVjOOQlC 80 (2.) The antwer to Uiii part of the qnestaon is 9 V 8 24 64/ 72^ '72 (8.) If AB be the rod ; P, Q the points of divuion, . x — 3 ^ P nearer A; AP « x, AQ - Y; then let no find the ^ chance that af»+(y-a?)»+(a-.y)a < f!, ^ ^c where x>{S<ay y>x<a. Take axes Ox, Oy inclined at 120°; measure OA « OB « a, and complete the rhombus OACB 5 o^ b sS then to every possible division of the rod corresponds a point in the triangle OAC, and to every favourable divimon a point within the circle a'+y'— ay-ay+ ^ = ^; 2 2m or, transferring to the centre X«-XY + Y»- |! - —. When n^2, this is the inscribed drde ; and when » >2, the circle lies alto- gether within the triangle^ and its area : area of inscribed drde » : ~ « 2 (3— ») : n. Mod mnd area of inscribed drde : area of triangle » «• : 3 y^S; therefore chance we are seeking is — ^ ^ ^ ^ if » » -.; ^ 3nv^3 2 4ir + 3-/3 and since this is greater than 2, it is the value required. If fi<2, the chance that n times sum of squares is less than the square on whole is |z±(3.«2.+ 2,-6.). where «n.-(|z^)». The chance that 3 times squares on parts > 2 square on whole is ^— j-t 1 2ir „ 7 „ „ > 5 squareon whole is ^ + — -_. 2465. (Proposed by the Rev. R. Townsbnd, F.R.S.)— If a, 6, c be the three sides of a spherical triangle, and In the radius of its polar drde, prove the formula tan* k ■■ (<^ ^ cos c sec g - 1) (cos c cos g sec 6—1) (cos g cos 6 sec 0—1) 4 sin « sin («— 0) sin («— b) sin («— c) Digitized byVjOOQlC 81 SohUioH hy the FBOPOBBB. If |}^ g, r be the three perpendicalars of the triangle; ai and a^ hi and h^, 01 and e^ the three pairs of segmenfaB into which tiiey divide the three sides; Ai and Ag, Bi and B^ Ci and C2 the three pairs into which they divide the three angles; pi and p^ qi and 93, r| and rs the three pairs into which they divide each other; and h as above; then, rinoe tan pi tan p3 — tan 9i tan gs » tan ri tan rs a — tan^ *, and sinee, by Napier's rule, tan />tan/>i«ftancitanc3»tan6tanooo8A tan Pi a sin a^ tan B^ « tan /> cot B cot C sec a, sin 6 nn c cos A cos B cos C therefore, at once, tan' h » -^ smBsinC* cosacosicoso' which, substituting for the fnnctions of the angles their familiar values in terms of the sides, gives immediately the above. A process exactly similar leads to the corresponding formula for a plane triangle; viz., sin B sin C 82« («— a) («-6) («-c) [See Townsend's Modem Oeometry, Vol. I., art. 168.] 2877. (Proposed by Professor Stltxbtbb.)— If V ^yet+gtx+txy-k-wyg, Y '^fytt + ijtex+hixy + kafyg; prove that the resultant of u, «, XT> V is SohtHon ly the Fbopobxb. Suppose x+y^O, g^O, t^O; then all four equations are satisfied provided /-^«0. Hence (/-^) (/- A) (y-A) (/-*) (^- *) (k-k) is a fiwtor of tiie TCsnltant. Again, suppose x^y^^g'^—t; all four equations are satisfied provided f-^gwmh + k. Hence (/+^-A-*)(/+A-^-*)(y+A-/-*) is also a fitttor of the same. But furthermore^ it f-i-sf—h^k ^0, then at the pcnnt of intersection «»1, yai, i;»— 1, <«— 1, It may eaoly be verified that the planes u, XT TOL. XI. K Digitized by Google 82 reipectiyely touch the cubic sorfiMses o, V . Either of these fiicte warrants us in concluding that some power of (/•i-g—h—k) Mffher tha» iheJtrH must enter into the resultant Uenoe (/-^)(/-A)(/-*)(^-*)G^-*)(A-*) is contained in the resultant. But the order of the resultant in these letters is evidently 8*4-8, t.e. 12. Hence the above quantity is the complete re- sultant. I obtained the theorem originally by ordinary Algebra^ I forget how, pro- bably by a direct application of the dialvtic process; but the preceding method is more instructive, as embodying m a simple instance the transcen- dental law familiar to eliminationists, that a»y Mii$f«2ar% qfrelation between loci inUrtecting in virtue of a condition implies the appearance of a power, higher than the first, of the characteristic of that condition in the complete resultant. The theorem was wanted in an enquiry connected with the sub- ject of rectifiable compound logarithmic waves. A vast extension of it will be given in the answer to a subsequent question. NOTB Oir TBI -LLTR JUDOB HaBOBBATB'B SOLUTION OT THB QlTIKTIO. By the Rev. T. P. Kiekman, M.A., F.R.S. One of the final equations at page 94 of the Judge's j^thumous treatise is - S(A<). Ifso,itispo8dbletofind ai •»• jSi « Ai, a, •»- iS^ » A3, &c., such that where P* ■* yi, and Q* ■■ yj. As ^1 + ^s » 2 (51), p. 86, is a rational and symmetrical function of the 2% P* + Ci^ is one also; that is, P* is a two-valued function, whose values are P^ and Q*. It follows that P is also a two-valued function of the a^a, whose two values are P and Q; for if P had more than these two values, PPPPP + QQQQQ could not be a rational and symmetrical function. Hence P -f Q ob 2 ( AO is a symmetrical function of the «*s. That is, y| +y| - 2 (AQ = A2^ = A2 («-dO = 0, and yi«-y2» or ^i+ys « 2(51) « 0. This destroys the form of the conditioned quintic in g, at p. 82 ; for (51)i-0 gives, as (21) and (81) are each » 0, (p. 78). (p. 7), (51) -46i«- 5*154 + ^6-0, or bg is expressible in bi and ^4. Thus it seems that whatHABaBBATB has been solving is not the general quintic Digitized byVjOOQlC 83 2870. (Proposed by the Rev. J. Wolstbnholkb, M. A.)— QWen three p<nnt8 A, B, C, and a oonio; two points P, Q are taken on the conic such that the pencil A f BPQC j is harmonic ; prove that the envelope of PQ is a conks toaching AB, AC at points on the polar of A with respect to the given Solution hy Abohxb Staklet. Since the intersections x and af of every connector PQ and the given lines AB, AC are harmonic comugates with respect to P and Q, the polar of one mnst pass through the other ; in other words, x and x^ are conjugate points relative to the conic. Conversely, every connector of a pair of conjugate points Xf x^ on AB and AC, being harmonically divided by its intersections P, Q with the conic, must be a tangent of the required envelope. But to a point a; on AB there is but one conjugate af on AC, and vioe versd; hence x and a/ are corresponding points of two homographic rows, and their connector xj/ envelopes a conic which touches AB, AC at the points which correspond to A, that is to say, at the points B, C, where AB and AC are intersected by the polar of A. This envelope has obviously, in common with the given conic, the four tangents at the points where AB and AC cut that conic. The following are amongst the most interesting special cases : 1. If ABC be a self-conjugate triangle, the envelope degenerates to the point-pair (B, C). 2. The envelope also degenerates to a point-pair (A, A')» if A be on the given conic, and A' be the intersection of the tangents at the points where AB and AC cut the conic again. 8. If AB and AC touch the conic in B and C, the envelope will likewise touch it in these points. 4. If A be at the centre of the given conic, the envelope will be a con- centric conic having AB, AC for its asymptotes. If AB, AC be themselves asymptotes of the given conic, then the envelope will be a concentric, similar, and similarly placed conic. Many well known theorems in conies immediately follow from the above special cases, wherein, it should be observed, AB and AC may be imaginary. For instance, if these lines pass through the circular points at infinity, then the envelope under consideration will become that of chords of the given conic which subtend a right angle at A, and this envelope will have A for its focus. If A be itself a focus of the given conic, then the envelope (by 3) will have in common with it this focus and the corresponding directrix. If A be on the conic, however, then (by 2) the chords which subtend a right angle at A will all pass through a point A' on the normal at A. (Salmon's Conies, Art. 181, Ex. 2.) 2828. (Proposed by S. Tebat, B.A.)— A straight rod is divided into n parts in arithmetical progression, and equal partides are fixed at the points of division. If the system be made to vibrate about one extremity, determine the average length of the pendidum, neglecting the weight of the rod. Digitized byVjOOQlC 84 SolutUm by the Fboposxb. Let a be tbe length of the rod, L the length of the pendulnm for any adjustment of the particles ; «!» 14 ^n~l ^^^^ distances from the point of snspennoQ. Then i«i + «s + .... +«,_! " 36(«„)' Let Ui^ae, d^~«i ■> x-t-y, where y is the common difference. Therefore «„ «i»{2a?+(n-l)yj; + ,^n(n-l)(2n-l)(3w;»-3»-l)5f«, 3K) -- in(n-l)(2ar-y)+ An(ti-l)(2n-l)y. But lw{2af+(w-l)y} = a; therefore y « ^^?~^x » therefore 2(i«:) - g.g-^_|?^ «V+aii(ii-2)(ii+l)(3»+l)a; + 2a2(n-2)(3»'-6ii + l)J, S(«,) - ^n(n+l)a: + ia(n-2). Now the number of pendulums is proportional to ^ this being the greatest value of X, Therefore the average length of the pendulum is a J S(«^ 10»(»«-.1)( n + l ^2(11-2))' If » be increased indefinitely, this becomes If the distances of the particles irom the point of suspension be in arith* metical progression, and one particle be attached to the lower extremity of the rod, the average length of the pendulum is ^^j£-^{4(2»-r) + 2(» + l)log2}. 2533* (Proposed by Professor Hibst.) — Find the envelope of a line upon which two given conies intercept segments which have a common middle point ; and find also the locus of this middle point. Digitized byVjOOQlC 85 I. SoluHon hy the Rev. J. WaumasnoTXi, M.A. If the two conies be Ufi+my^ + nz^ ^0^ Z'^-i-m'i/' + nV ^0, and vx+tHf+wz =0 be a gtraight line divided as required^ we must have (m»'-m'«) JL , + (nl'^ n'l) JL + (lm'~ I'm) JS- = 0, r— to fo— « «— « the tangential equation of the envelope. For the locos of the middle pointy we have '*The envelope of a chord of a conic bisected by a given straight line is a parabola of which that straight line is a tangent." Hence the points in which a ^ven straight line wUl meet the locus are the points in which it will be met by the two remaining common tangents to two parabolas ; two points : or the locus is a conic. II. Solution by the Veopobsr, Let (C) and (Cf) be the two given oonics, L one of the lines whose envelope is required, and I the common middle point of the two segments which (C) and (CP) determine upon L. It is obvious that the intersections of (C) and (CQ with L determine an involution of which lis a double point, the other being at infinity ; and it is well known that every conic passing through the intersections of (C) and (C) will cut L in a pair of cox^jugate points of this involution, that is to say, will intercept on L a s^meut which has the same point I for its middle point. One of the conies of this pencil, in fact, will touch L at I, another will have L for its asymptote, and consequently have its centre on L. Now the locus of the centres of conies of the pencil b^g aconic (2), there will be another conic which has likewise its centareon L, and that cenlare must obviously be the common middle point I. Conversely, every point on (2) is a point I on some line L. For, besides being the centre of a conic of the pencil, it is the middle point of one chord of (G). Two conies of the pencil, therefore, intercept segments on this chord which have a common middle point I; hence all do so, and the chord is a podtion of L. The required locus of I, therefore, is the conic of centres 2. Moreover, the above investigation has shown that the required envelope is that of the asymptotes of the conies of the pencil [(0), (C)}* Hence its class can be easily shown to be 3. For through an arbitrary point x at in- finity only one conic of the pencil passes, and therefore only one proper asymptote. The pencil however contains two parabolas, so that the line at infinity must be counted twice as an asymptote through x. The required envelope being of the third class, and having a double tan- gent at infinity must, of course, be of the fourth order. 2835. (Proposed by T. Cottbbill, M.A.) — In a plane, if A, B are two p(nnts, and a point P describe a curve of the nth order, show that P', the mtersection of the perpendiculars in the triangle PAB, will describe a curve of the 2nth order with three multiple points of the order n. Explain why the curve corresponding to a circle through the points A and B includes its reflexion to the line AB. Digitized byVjOOQlC 86 SohOum hsf the Pboposbb. (L) JM AC, BC be the perpendicDlan to the lino AB at the points A, B. Then, if P be a pomt in the phme, not on the lines, and PA, PB be joined, the per- pendicnlars AX, BY on PB, PA will intersect in a point P' i the pair will conjugately correspond, since each point is derived ftom the other by the same con- struction. The pur lie on a line through C, which is made np of such pairs of points ; amongst the lines is the line at infimW. Let A, B, C be called principal points. Then, if P lie on the line AC, V coincides with A. Similarly, to a non-^ncipal pcnnt on BC the point B always correroonds. If P is a non-principal point on AB, the point C cor- responds. Thus to every non-principal point of the plane one and but one point corresponds, whilst to eadi of the principal points the points of a line or the line itself corresponds. To a curve, therefbre, of the »th order, which does not pass through a principal point, a proper curve of the 2»th order will correspond, having multiple points of the order n at the points A, B, C. The nature of the correspondence of such curves is more clearly seen by observing (2) that since the angles at X and Y are right angles, the points X and Y lie on the circle described on AB as a diameter, and that P, F* are harmonic conjugates to the circle. But the line PP' always passes through C, the pole t/t the diameter AB ; so that we have a particular case of quadric inversion, the general prindples of which have been fully expluned by Prof. HnuST in his Piper on Quadric Invernon contained in the Proeeedingt qfthe Boyal Society for March, 1865. Corresponding points are therefore inverse points. One peculiarity of this inversion is that a right hyperbola through A, B is its own inverse; and rince the pole of PP' to such a conic passing tluough PP' is on XY, it follows that tangents to a curve at P and its in- verse at V meet on XY. (8.) The algebraical equations of the correspondence are very simple. Let («> v» y)i («'* v't ^) be the perpendiculars from PP' on the lines AC, BC, and AB, the Hue PF cutting AB in M. Then PM . F'M » AM . MB, so that we have the equations vf^v, t/s «, and y[y » uo. Also y^ y y \ yy The equation to the drcle AXB, the locus of coinddent inverse points, is im-y3 1« K - 0. The right hyperbola (axis AB) is uv+y^ «= H « 0. If («, v,y) ==0 be the homogeneous equation to a curve, (uy, vy, uv) is its in- verse ; (t^, vj/, H) s is its own inverse ; the inverse of (icy, vt/, E) «* is its reflexion to the line AB, &c. In tradng a curve which has a rimple inverse, the (xy) system of coordi- nates u often convenient, in which x is the perpendicular from P on OC, O being the centre of AB, y remaining the same as before. If AB a 2a, the inverse coordinates are connected by the equations x*^x, y'y ^c^^o?. Thus the inverse of the drde (y + 5)' » a'»^, which touches the lines AC, BC, is (6y •»- o^'^ofif =. (a'— «*)y'» a quartic curve having cusps at A and B, and a double point at C. If the drde does not cut the line AB, the quartic, b^ its revolution round OC, wdl generate a surface in the shape of a bell with its mouth uppermost, containing a bowl which just fits it at the rim. The inverse of a circle through AB is its reflexion to the line AB, because any point and the reflexion of its inverse to the line AB are on a circle through AB, as is easily seen. Digitized by Google 87 282L (From Wolstekholmb's Book of MathemaHeal J^roblenu.)-^ Show that the mean value of the distance from one of the foci of all points within a given prolate spheroid is ^a(3 + 0^, 2a being the axis and e the eccentricity. Solution hy R. TuOKEB, M.A. Let P be an element of the ellipse at a distance r from the focos (S), and the angle ASP ~ 9 ; then, by the reyolation of the carve about AS, the element at P generates a small ring equal to 2iir sin 9 r A9 An The volume (V) generated by ASL « 4iro8(l-62)(i + tf)8(2-ff), andthevolnme ^^ (V) generated by A'SL « ^ira8(l-e»)(l-c)2(2 + tf); hence -^ ^ (2-e)(l^ey ..^ -ZL « (llfKlz^'. v+v ii ' vTv' 4. For the part to the right of the boun^ng plane through SL, putting ff » £lLll^, the average distance of P is l-.eco89 ^yy',*sm<..»<Jr-|(l±i)(8-8e+«») (,). o • o* The average distance for the poridon to the left is found by changing into -c; thus it equals % ("IZ-f \ (3 + 3e + c>) {B). 2 \2+ €/ The average distance required must be determined proportionally to the volumes ; thus it equals 2815. (Proposed by A. Mabtiv.) — ^A cask contains a gallons of wine. Through a hole in the top water or wine can be let in at the rate of 6 gallons per minute ; and through a pipe in the bottom, when open, the mixture can escape at the same rate. Suppose the discharge pipe is opened at the same instant that water is let in at the top, and t minutes afterwards the water is shut off and wine let in. Bequired the quantity of water in the cask at the end of ti minutes from the opening of the discharge pipe, and the length of time elapsed, both before and after wine was let in at the top, when the quantities of the two fluids in the cask were equal, supposing them to mingle perfectly. Soluiion by C* R: Rippik, M.A. ; R. J. Nslson, M.A. ; the Pbofobeb ; and others. Suppose both pipes to have been open for any time ti and let ^ be divided into 11 equal intervals, and conceive the alteration in the fluid to hare been Digitized byVjOOQlC 88 effiacM hj removing ■— gftllons in each intenral and replenishing it with — • n gaUona of water. Then, if V be the number of gaU. of wine at the beginning of an interval, and Y" be the number of gals, of wine at the end of the same interval, we have V"= V- — . — = V /"l - — "^ - Te (o being constant) ; an \ an/ wo that, if y be the quantity of wine at end of time t, then, nnca there were a gallons at first, we have v.«...(>-g)-. Hence, making n infinite, we have for the result in the proposed question, that the wine at end of time ^ » ac * , and therefore the water « a (1— c '). Denoting this by a\ we have, by the same process, that the water in the cask after a fhrther time ^ « o'c * » a'c ' ; and the wine and water -^ a a will be equal when at ^ ^ z ^^ ^ « r log 2 (1), and also when a'* ""^.i ®' ^'^h^'T (^)' 2816. (Proposed by the Editob.) — Find three square numbers in arith- metical progression, such that the square root of each (a) increased or else {0) diminished by unity shall give three rational squares. I. SoluHon by Sakitiel Bills. If we put a-2pg + (2?*-g'), J « 2pg— rp^— ^2), and o-p'+j', then will a', 2^, 0^ be three square numbers in arithmetical progression. Now put 2!, ^, ^ for the roots of the three squares required in the y* y* y* question ; then in the part (a) of the question we shall have to find ?? + 1, — + 1, — + 1, or CM? + y*, te+y", osr+y* all squares. y* y* y* Assume a»+^-(r+y)» (1), and 5d?+^-(«+^)' (2). From (1) and (2) we readily find X -^?^ -*y-^ Digitized byVjOOQlC 89 Substituting these values in cx+y^, it becomes as—br 4 (as — brY* which is required to be a square ; that is, we must have 4cr» (r— *) (fis^br) t (br^—eu^ '^ a square. The above reduces to 4V— 4Aci^s + (4ac + 4Ac — 2ab) r*«*— 4aer^ + a V « a square. Assume the above - J^r8-2cr»+ f^^-^+ 2c-a^ jS^. Squaring this expression and comparing it with the above, we have -iacr+aU - -4c (^^J^+ 2c-a) r + (?f£^+ 2c -<.)'.. By reducing this, we find r « ^"^ "^ J. We give the following numerical examples, remarking that p and gr must be so assumed as to make s positive. o 1. Take p^2 and g»l: then a»7, b^l, <;a5; whence r ».#. 2 6 19 Take #=2, then rs3; whence d? b — , and y «« . 11 22 This gives ?5i 1??2, 1?*2 for the roots of the required squares. 2. As another example, take p^l and 9f'*>4 ; then we get a«89, &b23, c-665 which gives r«lZ». Take *=46, then r=47, a?- -?i, y « -5?^ 5. 46 131 262 whence we obtain for the three roots, }^^, 55215^, !*?§??»*. (5979)3 (5979)2* (5979)« 3. For a third example, take p-^5, and q^2; this gives a » 41, 5 »< — 1, e»29; or taking b positive, as we are at liberty to do, we shall have d»41^ -|Q A*-l, c«29 ; whence we find r^ — *. If now we take *«2, we have r»]3s 2 whence « - ??5, y = A, ± » ?????. From these values we find the 69 ^ 138* i/« 25 rooU of the three squares to be 3157*44, 91565*76, 12945504. Another solution may be obtdned as follows : — Put a-2pg + (p2-gr«), 6«p2+28, c = 2pg -(;)»- g^) ,' then will a^, b\ c^ be three square numbers in arithmetical progression. Now let ff , —f £^ be the roots of the three squares required in the question ; y» yi ya then we have to determine x andy so that ax + tf'^, bx+y^, cx+t/' may all be squares. Assume oay+y* =* P, bx+y^ « m', ca?+y* = »'; then ?nf_!!fn?_:^Zy! -«,(-«< suppose); a be therefore P— y' — ffr«f, m^— y' « ftra/, i|3— y^ = cr*<. TOL. XI. L Digitized byVjOOQlC 90 AMome l-i-y-iar, I-y-t^ «i+y-*t, m-y-rt, n+y-ee, ti-y-r*. We matt now have S^ -■ at^gt -• hs^rl — c<— r*. wbenoe (<+A)j» + (a5-<»)*-c/(< + a) - 0, therefore (^- o^)' •«• 4e< (^ -f- a) (< + &) must be a square. This redaces to l« •!• 4c<* + 2 (2ac 4- 26e— o^) (> -I- 4a&c^ + a^i* ; whieh aamime ■■ (l> + 2of + 2ac + 2^— 2c*— oft)' ; then, by redocixig this, we find t^ \ (c— 6— a). By symmetry, we have * • J (6 — c — a), r *- J (a— 6 - c). Also^ y - 4(ar-*0 - i(a' + ^+c»-2a4-2ac-2ftc)i wbfinfiA f. . 8( a-hc-&)(a + ^"C)(a-5-c) wii«uoe ^^ (2aA + 2ac + 2Ac-a«-6«-c8)a ' Therefore the roots of the three required squares will be 8a(a+c— 5)(g'|-6-c)(a— 6"c) 8/i(g-|-c— 6)(a + 6— c)(a— ft-c) (2a^ + 2ac + 2&c-o«-6«-c»)a ' (2a6 + 2ac + 26c-a«-6»-c»)» ' 8c (g +c~ft) (g-l- &— c) (g ^ft— c) (2aJ + 2flk? + 26c— aa-6»-cS)« ' where a, 5, c must, to satisfy (a), be the roots of three square numbers in arithmetical progression, such that, taking them all positively, the greatest root must exceed the sum of the other two. The results g^ven in the former solution may be readily obtained from this. If the greatest root be less than the sum of the other two, the result would satisfy (^), that is, it would give three square numbers in arithme- tical progresnon, such that, if each of their square roots be taken from unity, the three remainders shall be rational squares. For example, take j>— 8 and g»2; then a»17, ft»13, o»7; and we find for the three roots ax _ 108224 bx ^ 78936 c« ^ 42604 y* " (855)3 ' y* " (365)*' y* "^ "" (355)«' Other answers may be readily found. II. Solution by the Frofobeb. Let x,y,z be the roots of the three required square numbers, then these . have to satisfy the conditions aj»+«8-2y«, « + l-m«, y + l^n^, z + lm^p^ (1,2,8,4), where m,n,p are rational numbers. From (I), by the aid of (2), (3), (4), wehave 111? . 4^:1?, or ^^'^"'-Z ^ (it-Hp)(i>-y) ^^^ y + z x-y n2+p2«2 (»+«)(»»-») Assume the relations m + n ^r(n+p), m—nsss(n — /;) ; then from these we obtain 2L_«_!L.«_P_('»i suppose") (6). r-» + 2r* r+» 2-r + * V * / Digitized byVjOOQlC 91 Substitttting these vfilues in (5)» it becomes ;t5»(r-*+r»+l)« + (r*+l)2-H4i!Zi (T); and we have to determine r and s so that the second side of (7) may be a perfect square. In order to effect this, let r«— 1 «=/; then, patting for shortness' sake sf — ^ ff+ 2 + - j , (7) becomes h^^^r+sy-hi^ir-s+f) [=(r + *-25Fj8 suppose] (8). Prom (8) we find r = J(y-/) ; and thence, by the aid of (6), (2), (3), (4), we obtain the following general expressions for x, y, z, the roots of the re- quired squares : 4(r< + r-y)(rg-g+gf) 4y(r + g-^) 40-r + g) (l + g-.y) where g = i/+l+ :?* ^ '^ H£f-f)> and * = —t/, / being any rational number whatever. Taking /» 1, we obtun the numbers given as the results to satisfy the first conmtion (a) when the question was originally proposed, viz. . 264 1320 1848 86l' 861 ' 861* Taking/=i, we find _ 42504 78936 _ 103224 , "* ~ (355)*' ^ " p5j^' "* " (365)3 ' hence these fractions are the roots of three square numbers in arithmetical progression which satisfy the second condition (jB) in the question, that is, they are such that, if the roots be taken from unity, the three remainders will be rational squares. 2486. (Proposed by G. O. Haklow.)— Find a point on an ellipse such that the normal produced to meet the curve again may cut off a maximum area. I. Solution by Pbote&sob Whitwobth. Let PG be the required normal meeting the conic again in G. Let F be a point on the curve adjacent to P, and P'G' the normal chord at F. Let O be the point of intersection of these two normals. Then ultimately, when F' coincides with P, O is the centre of curvature at P. * Since PG is the normal which divides the area the "^c^^ most unequally, therefore the greater area cut off by PQ is a maximum. Hence ultimately area PP'O » area GG'O. But these areas are ultimately in the ratio of the rectangles OP . OF and OG . 0G^ Digitized byVjOOQlC 92 Therefore OP. OF* 00.00' ultimately. But if Cp.Cpf be semi-dia- metert parallel to OP and OP, we have OP.OG : Or.OG'- (y : Cp'*; therefore 0P« : OG^ - Cp» : Op", or OP : OG'- Cp : €/>•. But nltimatelj Co « Cp'; therefore ultimately OP » 00" or OP » OG. Henoe PG is the diameter of curvatare and the common chord of the ellipee and circle of curvature. Hence PG and the tangent at P make equal angles with the axis, viz., angles of 46^. Therefore the normal required is a normal inclined at an angle of 46® to the axis. OoBOUiBT. — ^The normal which cute off the smallest dosed area from any oonio b inclined at an angle of 46® to the axes. [A solution by Mr. Jimcurs has been given on p. 66 of Vol. X. of the II. SotutHmhyHeEDJTOig. Let C be the centre of the ellipee; PCQ the diameter, and PGH the normal chord through P; NPUand MHV the ordinates at P and H, meeting the circle circumscribing the ellipse in U and V; and UCW a diameter of the circle through U. Then it may be readily shown b\ that U, G, V are in the same straight line ; and by a well known property we have area AGP AGH APH ' AGV "* AUV PHQ b ■ — — ^ — f a areaAGU UVW a and b being the semi-axes of the ellipse. PHQ, and p for the angle VOW, we have Hence, putting S for the area of S - area PHQ . 1 (UVW) - ^oi (^ + sin ^). Now H«?, and PN b' CN GN b«' hence tan UGN » ^ tan 9, b and tanUCN = -tnn«; a therefore tan i^ » tan CUG » tan (UGN-UCN) : a2-6» 2ab sin 2a. Hence S » ^oft (^ + sin ^), where 2a6 tan ^ = (a^— 6>) sin 29. When S is a maximum, we must have whence we find the angle PGN « ss 450. Digitized by Google 93 Hence we see that the normal cbord PGH, which makes an angle of 45^ with the major axis, divides the ellipse into two segments, one of which (APH) is a minimum, and the other (BPH) a maximum. 2527. (Proposed by J. Wilson.)— Find the locus of the intersection of two tangents to a circle, the chord of contact of which subtends a right angle at a fix^ point ; and also of the middle'^points of a system of chords of a circle, which subtend a right angle at a fixed point ; and thence show that the envelope of a system of chords of a conic, which subtend a right angle at the focus, is another conic of coincident focus ; and that the locus of the middle points of a system of chords of a conic which subtend a right angle at a fixed point is another conic. Solution by the Bev. J. Wolstbkholmb, M.A. Takinff the fixed point for origin, let equation of circle be (a?— c)"+^«a', and let & + my «« 1 be a chord. The lines joining the points of intersection to the origin have for their equation «* + y' — 2ca: (to + wiy) + (c" - a^) (te + my)* « J and if they be at right angles, 2-2ci + (c«-a«) (P + m«) - 0. We may then write the equation of two parallel straight lines satisfying the condition in the form 2(te + my)a-2c«(te+ii^) + (c»-a«)(P+ifi») - 0, and the envelope is therefore (2*>-2ca?+c»-a?»)(2i,2 + c«-a»)-(2ay-cy)«, or (j£=^ + -^ i . a conic whose foci are the origin and the centre of the circle. The middle point of the chord being the foot of the perpendicular from the centre, lies on the auxiliary circle of this ellipse, t. e. on the circle {2x — c)' + %' n 2a^— c*. The locus of the intersection of tangents at the extremities of such chords is the reciprocal polar of this conic with respect to a circle whose centre is a focus, and is therefore a circle. Of course the envelope of a system of chords of a conic subtending a right angle at any point is a conic with that point as focus; as we get at once by redprocating the property of the director circle of a conic If the equation of any conic be aa5'+6y'-*-'2A* + 2Ay + l « 0, and the straight Ime to + my » 1 subtend a right angle at the origin, a + h + 2(hl + km) + ^ + ni?-mi0, and the coordinates of the middle point are X ^ Y 1 IJi + mY kml-hm^ + 6l Mm-kP + am ^ bi^i-um^ " hl^ + cuH^ gmX-ftnr ■(am» + b^)(lr/-A«/ or «X + i»iY«l, Z(lr + iY) = m(.i + aX)j therefore - = *^ « ^ ; h + aX k^bY aX^ + l\i + hX + kY* and the locus of the middle points is apparently a curve of the fourth degree. . Digitized by LjOOQIC 94 2769. (PropoBed by A. Mabtin.)— From a point, taken at random within a triangle, perpendiculars are drawn on the sides ; find the probability that the triangle formed by joining the feet of these perpen^cnlars is acot^ angled. Solution hf the Rev. J. WOLSTXKHOLICB, M.A. In an acnte- angled triangle ABC, if O be the centre of the drcmnscribed drdes, and arcs of circles be described on each side touching the radii from O to the ends of the side, the points mnst lie without each of the arcs. The result does not seem to be very neat, and it becomes more complicated for an obtuse-angled triangle. (Proposed by W. S. Bttbnsidb, M.A.) — GKve a geometrical interpretation for the relation between the invariauta of two conies, viz., SolmUon by the Rev. J. Wolbtbnholmi, M.A. The interpretation is given in Salmon's Conie Sections, It is the condition that the 8 points of contact of the common tangents to the two conies may lie on two straight lines ; in which ease also the 8 tangents to the two conies at their common points will pass through two points, four (two from each conic) through one point and four through the other. An example is any ellipse and a concentric rectangular hyperbola passing through the foci of the ellipse ; but the common tangents are unreal. Another example of two circles of radii 1, 7, with distances between centres 10, has the common tangents real, but the common points impossible. 2109. (Proposed by Capt. Claskx, R.E., F.R.S.)— Two lines are drawn at random across a convex closed curve; determine the chance of their inter- secting. Solution by the Pkopobbb. Let P be a point within the curve, PQ a fixed line. Let AB be any line crossing the curve, the normal to which, Pp, makes an angle 9 with PQ, and let the in- tercepted chord AB ^ u. Let a second line, whose normal makes an angle d', be drawn intersecting AB ; the number of such lines is +« rin (0— 6^ according as is greater or less than 6': the values of 0, ^ to range from to V. The number of lines then that intersect AB ia •t /**Bin(a-l^da'+ n /''sin(0'- «)<W= u(l-co8#) + i»(I + coBe), Digitized byVjOOQlC 95 or 2t(. If Fp^Pt the number of lines intersecting all linee paraUel to AB is 2 rudp » 2A, where A is the area of the curve ; and consequently the total number of pairs of lines that intersect is 2irA. It remains to find the total number of lines that cross the curve. Let Pf p' he the perpendiculars from P upon a pair of parallel tangents to the curve, then p + p^ is the number of lines which cross parallel to those tangents, and the total number is therefore /■ (p+p')d9^V, where P is the perimeter of the curve. The required probability of inter- section is, therefore, l5^. pa 2773. (Proposed by Matthew Collins, B.A.)— When a given angle rolls upon a fixed parabola, the locus of its vertex is well known to be a hyperbola having the same focus and directrix as the parabola. Conversely, when a given parabola rolls within a fixed angle, show that its focus and vertex describe lines of the fourth and sixth orders respectively; find the actual equations of both these curves, and thence show that the vertex of the fixed angle is a conjugate point on the former curve, and that the equation of the hitter, when the given angle is a right angle, becomes the sides of the given angle being the axes, and a equal to the distance of the focus from the vertex. Solution bjf the Rev. J. Wolstbkholme, M.A. The equation of a parabola touching the axes of x, y, which are inclined at angle «, bemg f-\ + f- ) » h the coordinates of the focus are X Y hk * ■ A " h^ + k» + 2hkco8w and the equation of the directrix is jp (A + it; cos «) •!■ y (A -f A cos «) « Ait? cos «. Hence, if 4»% be the latus rectum, we have 2a- ~ h^-*'k^ + 2hkcoBw " [ ( A + ifc cos w)' + ( A + A cos w)* — 2 (A + ifc cos w) (A + A cos w) cos w j * 2A9A»BiM«« (AH^' + SAitcoswy y Digitized byVjOOQlC 96 whence the equation of the locus of the focus is a»(X» + Y»+2XY coe«> - X«Y«8iD««, a corre of the fonrth degree having a conjugate circalar point at the origint and having the fonr asymptotes X ^ +aoosecw, Y "- +aoosec«. The vertex is most readily found from being the point of contact of a tangent parallel to the directrix, which gives at once, if (X, Y) be the vertex, (kY)^ ^ (AX)* hi a (A» + ife» + 2Aifecosft>)* . h + kcoBtt ^ k + hcoBt0 '^ J?TW+2hFcoB0 " Bin2«» 7ik three equations from which to eliminate h, k^ In the case where « b iv, we have or ««-X«r»(X« + Y« + 8a»). 2774. (Proposed by J. J. Waleeb, M.A.) — A central conic (i^o^+a^^ » a^l^ is turned in its own plane about its centre through a right angle ; prove that the locus of the intersection of the normal at any point on the given conic with the same line in its new position is one of the two sextic curves (a:« + y«)« {«*(*±y)»±6»(a?+y)«} « («^+5»)a(«»-yV. Quaternion SohOion hy W. H. Laybbtt, B.A. The vector equations to the two normals are f» » a COST 4- ^dnT + jp(J'o cosr + a^^ sinr), and p'^ ^fiooar a sinr + |^(&ai8 cost— odasmr); o a and that these may coincide, we have —«. 0^(0*^1* T + b'oori'T) « a&+(^— a')sinT cost. Whence, if a< sin' r + b^ cos' t » P, the equation to the locus becomes ab.V.p =■ gin T. COST (a*— 6^ ^b.a(aanT + ftcosT) + a.^(aanT— &cobt)J« To transform this to a Cartesian equation, we have P. a? arinT.008T(a>— b^(asinT + 6c()ST), P.y = sinT.oosT(a'— &3}(asin T— bcosT); therefore P' {a«(a?-y)«+ b«(a?+y)»} «-4a«b«sin«T cos^t (a»-6^, (»« + ySjS pa « 4 sin* t cos* t (a«-d«)*, ' (ajs«y»)3p4 « i6a262 8in«T coe«T(a3-^*, whence the required equation. Similarly for the hyperbola. Digitized byVjOOQlC 97 2244. (Propoeed by the late W. Lba.)— Form 11 symbols into sets, 6 symbols in a set, so that every combination of 4 symbols shall appear once in the sets. 2712* (Proposed by the late W. Lea.) — Form 15 symbols into sets, 6 symbols in each set, so that every combination of 4 synibols shall appear once in the sets. 2755. (Proposed by the late W. LEA.)^Form 16 symbols into sets, 4 symbols in each set, so that every triad in the symbols shall appear once in the sets. Solution by the Rev. T. P. Eisexak, M.A., F.R.S. The problem of 4-plefc8 exbansting the triads of 16 is a case of the more ^neral theorem which I gave in the Cambridge and Dublin Mathematical Journal, 1853, p. 42, thus : " 2** ypung ladies can all walk t(^ether in fourp day by day till every three have walked together." The proof is very simple. If n»2m, we take for our 4* symbols the repeating variations that can be made of them in 4-plets. .Then every triplet is completed into a 4.plet by the rule that no 4-plet shall have only three rtb places either like or unlike. Thus the triplet abbadd . bbcaab . cdaaaa (m=6) is completed into a 4-plet of axes by the six dddadc. When nf^Zm + l, we join to the 4*" symbols made with abed the 4"* made with abed, and add this rule, that no 4-plet shall have pnly three Italic or only three Roman symbols ; thus, if out of 612 young ladies the three (abbd), (ahhb), and (adaa) choose to walk together on any day, their companion must be (adac), and I have shown that any triplet whatever will determine the entire arrangement for the day. To form 5-plets with eleven elements so as once to exhaust the 4-plets (Question 2244), we first form the 12 5-plets that contain the two elements Oa; thus Oal23 Oal47 Oal59 0al68 0(7456 Oa258 0a267 0a249 (Oa) Oa789 0a369 0a348 0a357 which are obtained by properly reading any one of the four squares of triplets in figures. We call this set of 12 (Oa). There must be in the solution a set of 12 (01), another (02), &c., another (12), &c., such that, if we form them all, we shall construct all the 11.6 quintuplets 10 times. Denote by (al, 4678) (Oa) the operation on (Oa) with the two circles al, 4678, that is, the exchange of a and 1, and the putting 6 for 4, 7 for 6, 8 for 7, and 4 for 8. Then we find that (01) = (al, 4678) (Oa), (al) - (01, 8764) (Oa), (02) « (a2, 5489) (Oa), (a2) = (02. 9846) (Oa), (03) -= (a3, 6697) (Oa), (a3^ = (03, 7956) (Oa), (07) = (a7, 1345) (Oa), (a4) « (04, 2197) rOa). (08) « (a8, 2166) (Oa), (a6) = (06, 3278) (Oa), (09) « (a9, 3264) (Oa), (a6) - (06, 1389) (Oa), (04) « (a4, 7912) (Oa), (a7) « (07, 6431) (Oa), (05) = (a5, 8723) (Oa), (a8) « (08, 6512) (Oa), (06) = (aJd, 9831) (Oa), (a9) « (09, 4623) (Oa). We have now every quintuplet containing and all containing a, in num. ber 12 + 18 + 18 = 48. If a - (4678), e-» = fl8 « (8764), and O^ « (47, 68) ; and the operation Digitized byLnOOQlC 98 (Oa, 47« 68) on (01) will give (al), and tfiee vend, (10) and (la) can be formed by readixig horizontally, yertically, and parallel to the dia^nals, th0 two triplet squares in the right members following; (10) » 10 . a23 (la) » la . 023 657 854 849 679 and we obtain la, 12, 18, &c from (10), thns (la) - (Oa. 47, 68) (10), (16) - (06, 29, 8a) (10), (12) » (02, 96, 54) (10), (15) « (05, 38, 42) (10), (13) - (03, 85, 79) (10), (IT) « (07, a4. 93) (la). B3,a6)aO), ,7a,25)aO), ,62, 37) (10). We hare now every quintuplet contuning 1. Next we writ^ by (2a) « (Oa, 94, 85) (20), (20) -20. lad (2a) » 2a. 103 846 596 795 748 and we obtain, omitting (2a) and (21) ahready found, (23) = (08, 56, 47) (20), (27) - (07, 81, 84) (28) - (08, 17, 5a) (20), (29) - (09, 4a, 16) (24) = (04, a9, 73) (20), (25) » (05, 63, a8) (26) = (06, 35, 91) (20), This gives us every 5-plet containing 2. Next, we write (30) and (3a) » (Ofl,69.57)(30), (30) « 30.12a 495 687 The process is continued thus : — (8a) -8a. 120 467 985 - (06, 9a, 14) (80), (08. 61, 29) (30), (07, 42, a6) (30), which gives all containing 3. We now add the 54 following to the 12 above formed with Oa : — 01657 02795 al854 a2748 12947 17958 01849 02347 al679 a2397 12856 23875 01254 02365 al257 a2368 12376 23496 01379 02389 al349 a2345 12348 24765 01358 03495 al356 a3467 12359 24589 01629 08687 al829 a3958 18547 26789 01278 04587 al246 a4759 13689 34789 01346 04679 al378 a4986 14569 34568 02846 05689 a2596 a5876 14867 35679 If it is possible to form quintuplets with 15 letters so as to exhaust the quadruplets, there must be 26 containing the final letters de. These are easily formed by cyclically permuting the 18 elements 123 .... 90a6o under 125 and 138. This gives us {de)^del26, d^2S6, .. ..c2el38, ({e249 There must be a set (di) of 26, and another (el). We readily find (le) = (Id, 2358, 7a90) (de), (1<0 « (le, 8532, 09a7) (de), results so fiir analogous to the preceding. Digitized by Google 99 Bat I cannot find the subjtitacions that will correctly give (2«), (2<i), &c. I shall be a^preeably Borprised to see this problem solved. It woatd be plea- sant to find it to be a simple evolution of the iamoiis fifteen yonng ladies whom I had the honour of introducing to the planet for the first time in the LadAfa and GentUmafCa Diary for 1850. To my original puzzle of seven arrangements of five triads so that every pair should once walk abreast. Professor Stlybstbb added a second, to make the school of 15 walk out every day in the quarter in 13 . 7 columns of triplets, till every three have once walked abreast. I gave, at the time, one solution of Ph)f. Stltbsteb'b question in the Cambridge and Dublin Mathematical Journal, It would be a memorable sight to see his 18 . 7 columns of 6 triplets made to perform the right-fiice, so as to present the 13 . 7 . 8 quintuplets here required. Any field-marshal could transform by a word the column of 5 . 3 into a dangerous Ehalanx of 3 .5; but I fear that a couple of generalissimos would find it a eavy strain on their tactics to turn out day by day for the quarter the exact quintuplets of Mr. Lba. 2893. (Proposed by M. W. Cboittok, F.B.S.)— If there be {n) quantities a, 5, c, d, oc, each of which takes independently a given number of values aia^a^ .... bih^h^ . . . . &c. (the number may be different for each), if we put 2(a) « a + 6 + c+rf + &c., and if for shortness we denote " the mean value of «" by M (;r), prove that M [2(a)] = M (a) + M(b) + M (c) + &c. « 2[M (a)], M C2(a)]2= {2 [M (a)]}'- 2 [M (a)]* + 2 [M (a«)]. Sohdion hy W. H. H. Hudson, M.A. Here a may have any of r values, ai^ oj, og ... ; b any of # values, hn &2, b^ ... ; e tjoj of i values, 0|, c^ e^ ... ; &o. &c. ; therefore a + 5-ho+ ... may have any of the rst... values ai+6j + ei+ ..., Og + Aj+Cj-f- ..., , whereof ai occurs in «^... sets, bi in rt„. sets, and so on; therefoi-e ^(^^^•^^)" (- M2a) - *<»-»(^-»-«a + »>0 + <>«-(^ + ^-*- >♦■)•^■.. rst„. ' rai,,, ^ ai + c^+^ ^ Vijvt:^. ^ ^^^ ^ Ma+M6 + Mc + ... - 2Ma. r 9 Agfdn, (2a)^ may have rst.., values, of which the type is ...a? + 62+c»+... + 2aJ + 25o+... of will occur in H.,., b^ in rt,,,, and so on; ai^iwill occur in ^..., hiCi in r..., and so on Therefore M (2a)s « ^...(ai» + Oa' + ...)+r<...(6i»+...) + 2<...(ai5i + ...) + 2r...(&iCi+...) ret ^ ai« + Oa«+... ^ bf+bf+„. ^ ^^^ ^ Haih + aA+'") ^ 2(Vi+»-) . r • "** rs . 9t Digitized byVjOOQlC 100 Now ^ + -I- • . ..« r« st tberefm M(^)2 » M(aa) + M(&3)+ .... _ / fli + aa+... y_ / b, + b,+ ... y_ - 2Ma»+(5Ma)2-2(Ma)2. 2887. (Proposed by Asohes Stanley.)— !• If a point P on a conic be eouuected with any two fixed points A and B in its plane, all chords which are divided hflrmonically by PA and VB will be concurrent. 2. The locus of the point of concurrence when P is variable is another conic which has double contact with the given one on AB. Seiution hy F. D. Thomson, M.A. , 1. Let QB be the diord divided harmoni- cally by pa; PB. Then, constructing ns in the figure, -1 = p{3QTR} =:i p{cqdr} arR{CEDF}; therefore F is the pole of QR, and therefore QR passes through the pole of CD, i,e, through a fixed point. 2. If O be the pole of CD, then when P "varies the locus of O is a conic. For consider the points in which the locus of O is cut by the tangent to the given conic through A. Let the tangent through A have L for its point of contact. Let BL meet the conic in M, AM meet the conic in N. Then the poles of LM and LN are the points on the required locus upon the tangent AL. Hence the locus is a conic. Also the tangents to the given conic at the points where AB meets it are seen to be tangents to the required locus. Hence the locus is a conic having double contact with the given conic at two points on AB. . The particular case when A and B are the circular points is given by Salmon, p. 257. Digitized by Google 101 2728. <PropoKd by W. S. MoCat, B.A.)— Oiven Ibree planes and- tbdr poles with regard to a system of qnadrics, the locos of centre b a righ^ line. Solution hfi the Pboposbb. For the lines joining the vertices of a tetrahedron to the corresponding vertices 6f its polar tetrahedron belong to the same system of generators of an hyperboloid of one sheet. (Salmon's Geometry of Three Dimensiane, p. 179.) Hence, the centre being the pole of the plane at infinity, we see that - if through each of the given poles we draw lines parallel to the intersection of the other two planes, the hyperboloid of which these are generators passes through the intersection of the three planes, and the lecas of centre is the generator of the same system at that, point. 2597. (Proposed by W. H. H. Hudson, M.A.)— A right cone, whose weight may be neglected, is suspended from a point in its rim ; it contains as much fluid as it can : show that the whole pressure upon its surface is , ,. sin a cos 9 (cos (9 + a) /4 where h and 2a are the height and vertical angle of the cone, and is deter- mined from 8 sin Si0 « 4 sin 2 {&— a). Solution hy Stephen Watson. Let C be the point of suspension; ABC a vertical section of the cone; BF a horizontal line cutting AC in D, and meeting a perpendicular from A in F; and O the middle of BD. Join aO, and take OG » ^OA ; then G is the centre of gravity of the fluid ; hence CO must cut OF perpendicularly in E, so that OE— ^F. Put ZCBD — a, the area of the horizontal surface of the fluid «s S, and the area of the surface of the cone in contact with, the fluid » (A). Now BE » BC cos 9 «- 8A tan a cos a, BF -> BA cos (^— a-a) » A sec a sin (0 + a), and BO^-iBD-^^^^'*^^^'- *«^°« 2sinBDC cos(0-a) hence, by substitution in BF-BO -4(BE-.B0) or BF + 3BO-4BB, we have sin (0 + a) cos (0-a) + 8 sin a cos a = 8 sin a cos ooS (0— a), .•. ^(8in20 + sin2a) + |sin2«-4sino[cos(20-o) + cosa} - 2 {sin 2a-sin 2(a-a)} +2sin 2a, therefore 3 nn 20 » 4 sin 2(0— a), the condition for finding 0. Again, if a, 6 (a « OB) be the semi-nxesi of the surface S, we have, by Digitized by Google 102 potlxDff |v--a-i-« for Ot md a for /S, in eqoafcioo (12), p. 895. VoL IL, of IHiviw emtioo of IT " ' " 8 Alao the distance of the centre of gravity of (A) from the horiioiital line BFia iAF-iBA8in(^*«-e)-iA8ec«cos(e+a) (1). Moreover, the orthogonal prqjedaons of the aoriaGef S and (A) upon the baw of the cone most hie equal, that ii^ (A)8in«-Soo6«, therefore (A)-^S; ana hence the whole pte w uie upon the oone% anrfoce ia 210& (IVopoied by W. K. CuwrovD, B.A.) — Required analogoea in Solid Oeometry to the following propontioua in Plane Gkometry : — ia.^ The perpendicnlara of a triangle meet in a point. b,) The middle points of the dtagomtls of a qnadrilateral are in one straight line. (0.) The drdes whose diameters are the diagonals of a qnadrilateral have a common radical axis. (d,) Every rectangular hyperbola circumscribing a triangle passes throngh the intersection of perpendiculars. (0.) Every rectangular hyperbola to which a triangle is self-conjugate passes through the centres of the four touching drdes. (/.) mn (A + B) B nn A cos B + cos A sin B. (ff.) The sum of the angles of a triangle «■ two right angles. (A.) In any triangle ^ - 5L? - ?L2. SoUfUon hy the Pbofobib. I have an analogue for eadi of the four (cQ, (0), (f\ (jf), and more than one for each of the others obtained by extensions of Mr. Qsbbb*8 methods. (c.) A straight line cuts the faces of a tetrahedron ABCD in a, h, c,di the spheres whose diameters are Aa, Bb, Cc, Dd, have a common radical axis. Hence the middle points of these four lines are in one plane. Let ft conicoid whose asymptotic cone has three generating lines at right angles be called a rectangular conicoid. (d), (e). Every rectangular conicoid drcumscribing a tetrahedron whose perpenoiculars meet in a point, passes through the point. And every rect- angular conicoid to which a tetrahedron is self-conjugate, passes through the centres of the dght touching spheres. Digitized byVjOOQlC 103 (/) oiitABC)-rin(ABD+BCD + CAD) » mn (BCD).C08 AJ) + mn (CAD). cos fib + sin (ABD) . cob ci>, where A, B, C, D are four tines in space, and on' ABC - 1, COB AB, 008 AC COS AB, 1,^ cos 1^ ooBAb, oosBC, 1, (j.) In the triangle case this should he written {BC)+(CA)+(AB)-0. The analogue is then ohvionsly (BCD)-(CDA) + (DAB)-(ABC) = 0, A, B, C, D heing any four planes. (A.) In any tetrahedron, AC.DB (dMfd a ._ Y^ ^ a ^ tt ^f! iv^n D, sinA&.sinbB V« cos A iabcd)^ where a, 5, c, rf are the feces, and cos^ A » 1, cos BC, cosBDI COS Sb, 1,_ cos CD cosBD, cos CD, 11 (AB, &c denoting angles hetween planes). 2661. (Proposed by W. S. Btoksidb, M.A.)— Determine the form of the solution of the differential equation 2/(x)g + 8/'(x)^+{rW±«'}u-0 a). Solution hif the Frotobxr. The differential equation /(')g+*/'<'>|±-^-« <">• i. reduced to the farm g±»V-0 ^J "^""^ *^/:^^ ** *^* independent vwUble j whence, taking the lower rign, Agun, ffiibrentaating the equation (2) with regard to a>, ud writing U for ^ and 2/(») fiwr /(»), we tod the pven equation (1), the (ohiUon of which is therefore U-(A) ri^y-^Uhe"^"^]. Digitized byVjOOQlC 104 8895. (Propoied by J. J. Wa£kxb, M.A.) — At any point on a cntped cnbic a tangent is drawn meeting the cubic in a second point ; from the first point a line is drawn touching the cnbic in a third point. Prove that the lines drawn from the cnsp to these points form with the cuspidal tangent a harmonic pencil. SoUaion hy W. H. H. Hudson, M Jk. Let the tangent at A meet the cubic in a, and from A draw AB tangent to the cubic. Let C be the cusp, CT the tangent thereat. It is required to prove that CB, Ca, CT, CA form an harmonic pencil. Take ABC as tri- angle of reference. The equation of the cubic 18 y{la + mpif + fu?fi *= 0. [This may be obtained by taking the complete equation of 10 terms, and simplifying it from the considerations (1) that the cubic passes throneh A, B, C, (2) AB touches at B, (3) any line through C meets it in two points, then (4) the two lines that meet it in three points at C are coinddent.] la+mfi ^ is the equation of CT, Py + nfi^O is the equation of the tangent at A (Aa), and it meets the cubic where 2la + mjS » 0, which is therefore the equation of CA. From the form of these equations we see that CB, Ca, CT, CA form an harmonic pencil. [Mr. Thomson and the Pbofosbb remark that the theorem follows at once from Salmon's Higher Plane Curves, Art. 178.] 2002. (Proposed by J. J. Waleeb, M.A.) — The circle aissing through three, points on a parabola, the normals at which co-intersecl]Mlways passes through the vertex ; and if the point of oo-intersection of the nomals describe a coaxial conic having the vertex as centre and axes in the ratio of m : n, the locus of the centre of the drde will be a conic having the focus as fMiitreb and axes in the ratio of 2m : ». Solution hy the Pbofobeb; W. Robbbts ; R. Tucebb, M.A. ; and other». Let (a, /3) be the common point of the normals, and y' — ^ax the para- bola ; then the equation to the drcle is readily shown to be aS+y»--(a+2a)a?~i/3y-.0 (1). From this equation wo see that the circle passes through the vertex; also that it U, f/) be the centre, 2*^ « o + 2a, 4^ = fi. Let n^a + m^0^ «= *» be the locus of (a, jS), then substituting 2(ir'— a) for a and 4/ for fi, » V* + 4m V^ » -^^ is the locus of (p/, yO^^ ^^''^c having its centre at {—0,0) and axes'in the ratio of 2m : ». Digitized byVjOOQlC 105 2698. (Proposed by the Rev. J. WoLSTsyHOLiCE, M.A.)^A circle rolls with internal contact on a fixed circle of half its radius; prove that the en- velope of any chord of the rolling circle is a circle which reduces to a poiut for a diameter. Solution hy W. H. H. Hudson, M.A.; P. D. Thomson, M.A. ; R. TuOKEB, M.A.; and others. Let the diameter considered be the line joining the centres when the point of contact is A; when B is the point of contact let D be the centre of the large circle. Join DA; and pro- duce it to meet the large circle in E. Since BOA = 20DA, arc BA = arc BE ; therefore E is the point to which A has rolled, and DE is the new position of AC. Hence the diameter always passes through A. Now conceive a chord parallel to this dia- meter. It is always at a constant distance from it. Hence, if from A we draw a perpen- . . dicular upon it, this perpendicular is constant. It therefore envelopes a circle with centre A, which degenerates to the point A when the chord becomes a diameter. 2875* (Proposed by the Rev. J. Wolstenholmb, M.A.)— Any tangent to a conic is^ of course, divided in involution by three other tangents, and the hues joining their points of intersection to one of the foci of the conic ; prove that the distance between the double points of the involution subtends a right angle at the focus of the conic. The locus of these double points, the three tangents being fixed, is a cubic having a double point at the focus, whose nature I have not examined. Solution hy Abches Stanley. Let A, B, C be fixed tangents to the given conic, an^X a viuriable one; moreover, let P and P' be the tangents from a given point p, (In the ques- tion p is a focus, P and P^ therefore pass through the circular points at iur finity, and all harmonic conjugates relative to them are orthogonal.) Now, by a well known theorem, the rays from p to the three pairs of opposite intersections of the quadrilateral ABCX are in involution, and to this invo- lution likewise belong all pairs of tangents from p to the several conic^ inscribed in the same quadrilateral. Now the given conic obviously belongs to this series, hence P, P' form a pair of conjugate rays of the involution; the double rays D^, Dj thereof must, consequently, be harmonic conjugates rela- tive to P and P', no matter what position X may have. This proves the first part of the theorem. ' With respect to the locus of the intersections of X with the double rays Dj and Dg, I observe that the latter are the tangents at p to the two conies inscribed in ABCX which pass through p. Hence, not only does X determine the harmonic conjugates Di, D2 relative to P and P', but, conversely, every vol. IX. N Digitized byVjOOQlC 106 raeh line-pair D^ Bj determinee its oorresponding tangent X ; the latter is, in fact, the fonrtib tangent common to the given conic, and to that whidi is inscribed in ABC so as to toach Di (or D^) at p. It is easy, therefore, to determine the order of the locns, for through any point | of an arbitrary line L will pass one line-pair (Dj, Dj) corresponding to one tangent X, which cuts D say in w. Again, two tangents X pass through every point » ; and to them will correspond two line-pairs (Di, Dj) cutting L in four points |. Hence, between tne points ( and x there is a (1, 4) correspondence, and con- sequently there are five points in which x and ( are coincident. It is easy to see, however, that one of the latter is always on P and another on P': de- ducting these, which are not proper points of the required locus, we conclude that the latter is a cubic, of which p is a double point and P, P' the tangents thereat^ since on every line through p there is but one point of the locus whidi is not coinddent therewith, and on P (and P^ there are none, (When p is a focus of the given conic, it is also a conjugate point on the cubic.) It IS easy to find the nine intersections of the cubic with the fixed tangents A, B, C. Three of them will be the points a,b,c, where the sides A, B, G are intersected by the connector of p with the respective opposite vertices (BC), (CA), (AB) ; for instance, a is obviously a double point of the involation on the second tangent X which passes through it. The remaining two points of the cubic on A may be best found by conceiving X to coindde with it, or, in other words, to intersect it in its point of contact a ; for then a and a will be one pair of conjugate points of the involution on A, another pair will be the intersections (AB), (AC). The points where A cuts the cu bic, therefore, will be those which divide both the segments aa and (AB) (AC) harmonically. In this manner the order of the locus might have been more readily deter- mined. 2003. (Proposed by W. H. Lavebtt.)— If A', B', C, IX, E', F be the minors which are the coefficients of A, B, C, D, E, F, respectively, in the different expansions of ' A F E I F B D J and if ^1 denote either (A', B', C, D', E'. P0(\, a, y)*, or E D C| (A',B',C,D',E',F)(a,6.c)»; if ^ t= (1, 1, 1, —COS A, —cos B, —cos C) (A.. /«, i^); ^-^ ^-($)(^)-(t)(^)' «<lQ'«'<lH«'-te»„>U., quantities involving y, \, and A, /», respectively; then show that, if a focus of the conic (A, B, C, D, E, F) (a, jS, 7)^ = be known to be in the line Aa+/iiS + iTr « 0, its coordinates {k\ /, •) are proportional to the minors of , and the equation to the diameter of the conic is the determinant \\ fi v IPQR a fi /d$i\ /d^\ /d^\ \daj \db ) \dc) = 0. Digitized byVjOOQlC 107 SoluHon hy the Pbopobbb. It is eyident that if we find the pole of Xa+ ftjS + ^7 » 0, and firom this pole draw a perpendicalar to the Une, this perpendicular will meet the line in the focus. The coordinates, then, of the pole of the line are _^ !L i- (1). m it) ©) and the condition that aline 'La+Tdfi + 'Sy »0 •• (2) may he perpendicular to Aft+ fifi + try ss 0^ we should find to he L(X-/iCoe C-y cos B) * MOi- fcos A— Xcos C) + K(y-Xcoe B-/&COS A)-bO, H^h^'ith'^it)-' <»>• Also we have, from (1), mnoe {a' fif '/)Ib on the line^ '■ityitym-' '*>' therefore, from (2), (3), (4), we get (by eliminating L, M, N) for the equation of the perpendicular Va+Qfi + B,y ^ (5), and for the intersection of this with ha+zifi + ry^ we have (for the co- ordinates of the focus) \' ji' • Iqr| |rp| and since the coordinates of the centre are P Q .„_J 7 (^\ (^\ (^\ \da) \db ) \do) we have, from (6) and (7), for the equation of the diameter .(6)S .•^ .P), V fi / y fd^\ fd^\ /d^\ \da) \db) \do) = 0. 2730. (Proposed by J» J. Wauebb, M.A)— If the determinant a e f a ^ * J ^ VMUA. prwe that (erf) - ^^m^MB, ^here (6). (0, g % h d (A), ipd), ... are minors formed by omitting the rows and oo^umna in which the bracketed constituents stand. Digitized by Google 108 Solution by the Pbopobsb. In the Itaungtr of Mathematict, No. XIII., p. 27, I, have sbown that a*\); cong«;qnently, when D— 0, \m (bd) h be) ^ ' (,bd)(bc)-ibky jei) {{e{)(bi)-{bJc)(dk)} ^■ (dh) [{d}i)(be)-{blc){ei)} ~ (bd) lbe)-Q)kf .(WKjl+mffl («»p.26). (6) which is the type of six similar relations. (Proposed by the Rev. J. Womtehholme, M.A.) — Four fixed tangeuts are drHwn to a couic S ; three other conies are drawn osculating S in any the same point P, and each passing through the ends of a diagonal of the circumscribed quadrilateral ; prove that the tangents to these cf>nic8 at . the ends of the diagonal meet in one point P\ and that the locus of P' is a curve of the sixth degree and fourth class, having two cusps on every diagonal, and touching S at the points of contact of the four tangeuts. Solution hy F. D. Thomson, M.A. With the letters of the figure take the triangle ABC d formed by the diagonals as the triangle of reference, and let the equation to the line DEF he ax + by-k-cz-0. Then the coordinates of the points in the figure may be taken as 5^^^^"'^* D (0, c, —M, G (0, c, i), ^z:-'n2^^c E(-c, 0, a), H(c, 0, a), F(6, -a,0), K(5,«,0). \ \ By properly determining the constants, we can take as ^n,\ the equation to the given conic B ar»+y2+aj2«0 (i). Since DEF touches this, we get the condition a2+62 + c2-0 (ii). But, by Salmon's Conies^ Art. 251, the equation to a conic osculating (i) at the point (a^i^ V) is a^ + i^^ + z^+ {xaf +yy' + zz') {lx + my + nz)^0 (iii), where 2a?' + wy' + fi«'= (iv). If (iii) pass through the points D and G, we find, substituting the values of the coordinates, u.y(6i^c2) ^ gy ^ _ ^aV Digitized by Google 109 and therefore I "(g^-.V^ . therefore (iii) becomes Now, if (a?y«) be the pole of DG, t. e, of jp=:0, {xt/z)iB given by £? « 0, ^ - 0. dy dz These equations, after reductions by means of (ii) and a:^+y^ + 2'2 = 0, or eflx b^ ^ ^j which, from its symmetrical form, is evidently the same point as we should have obtained from either of the other two osculating conies which pass through EH and FK. Equations (v) give us x^ if'^ z^ ' (a^a?)* (6V)' (c^^)* (a»af)'+&c.' . therefore the locus of the point is the curve ((Mr)* + (%)* + (c22)* = 0, or (a*xi + hy + C*z^f = 17a<bV«^V, a curve of the sixth order having two cusps on each diagonal, at the points where they are cut by the conic a^a^-i-by + c^z^ as 0. If we transform to tangential coordinates, the equation to the point given by(v)is ^K+^,i + ^y^O (vi); a* 0* c* therefore, to find the locus, differentiate, and we have. Z-Kda/^tliidy'^ L.ydz'^ 0, when af dx' -v if dtf ^ zfd^ ^ 0; a* 0* c* X, ^ af i/ ^ therefore — A.=s,^ttaa_-yj therefore, eliminating x\y\z\ ^ + — + ^ b (vii), A,» 11^ v^ the tangential equation to the locus. This is of the 4th class, and is satisfied by the lines ( +a, +&, -f c), and therefore touches the sides of the quadri- lateral. Also the pole 5 the line (x'^V) « XV (Mi/a + c*/»'*) + &c. « 0; therefore point of contact of (a, 6, c), t. e, of DEP, is aA + fe/* + ci' » 0, or in trilinears f^ + ^ « _, which is the point of contact of DEF with the given a b c conic ; therefore, kn. Cob.'- If the points F and E be the two real foci of the given conic, and E and H the circular points at infinity, B is the centre and DG the minor axis. Hence the osculating conic through EH and the point {afy'z') becomes the circle of curvature at afff9f\ that through DG and {p^y'v) an osculating conic through the imaginary fod; and that through FK and Qfjfi^ an oscu- Digitized byVjOOQlC 110 latinff oonio ihroagh the real Ibd. Henoe the locos of the centre of cnrvatnre at (Jyi^) ia the same as that of the poles of the minor and major axes of the original conic with respect to the other two osculating conies. 2787. (Proposed by C. W. Mebbifijexj), F.R.S.)<— The value of jfYTx'dxdyds taken over the whole volume of a tetrahedron throtkgh one of whose comers the plane of (yz) passes, is iV(ai' + 51^401*), where V is the volume, and ai, l^, ei are the distances of the middle points of the oppo- site edges from the plane of reference. SohOhn by the Pbofobxb. Let us first find the value of the integral taken over the volume of a pyramid having its ha§e in the plane of reference. Let the pyramid have a square base, and let one of its edges be perpendicular to its base. Let its height be h and the side of its base k. The Cartesian equation of one of its edges will be ^-j.^ k 1, and we shall have to find / sfif^dx subject to that equation. On substituting h(l—^ for y^ and performing the in* te^tions, we find its value to be -^ A^A' — -^ base x (altitude)^. It is evident that, whatever be the ^ape of the pyramid, we shall always have Jf/a^dxdydz » ^base x (altitude)*, provided the base of the pyramid be in the plane of reference. Moreover, the volume is \ base x (altitude). Now let ABCD be the tetrahedron, and DEFG the plane of reference. Produce the four planes of the pyramid to intersect with DEFG, as shown in the fig^ure, and let the distances of A, B, and C from the plane ot reference be a, jS, and y respectively. Since the pyramids ADFG and BDEF intersect so as to have the quadrangular pyramid DBCFG for a common frustum, we ^* ABCD - ADFG + CDEP- BDEG ; and also, if we call MI the integral Jfja^iadydn taken over the eorre* Bponding pyramid, we have MI . ABCD = MI . ADFG + MI . CDEF-MI . BDEG. But MI . ADFG « — DFG » ^ ADFG. 30 10 and similarly MI . CDEF . J^ CDEF, MI . BDEG = -^ BDEG. 10 10 Again, BDEG = ADFG + CDEF- ABCD; therefore XOMI.ABCD« (a«-i8«) ADFG + (r-/3^ CDEF + /8«. ABCD j Digitized byVjOOQlC - Ill next, AFGD : ABCD = A AFG : AABC » AF . AG : AC. A6 *=o.o: (a-7)(a-iB); also, CEFD : ABCD - aCEF : A ABC « FC . CE : AC . CB -7-7:(«-7)(^-7); . lOMI.ABCD a'(«^"ig^ . T^Ct'-ZB^) ^^ •• vol. ABCD (a-i8)(a-7) {y^a)(y-fi) ^ a— 7 0—7 a— 7 -» o»+ 132 + 73 + /37 + 70+oiB -i(« + i8)« + i(/8 + 7)3 + i(7 + a)«; therefore MI . ABCD « ^V (ai« + V + ci«). MI is not a moment of inertia, being taken with reference to a plane instead of a line; hot the formnla, as it involves the squares, evidently holds for the moment of inertia aboat an axis. The common moment about the plane is iV(ai + 5i + C|). Moreover, the common moment and moment of inertia of a triangle about a line passing through one of its angular points are, severally, A being the area of the triangle, J A (flfi + bi + cO and ^A (a^ + b|' + cfl. The mmilarity of these forms is very remarkable. 27(WL. (Proposed by the Rev. J. Wolstbkhoucb, M.A.)— In a right conoid whose axis is the axis of z, show that the radii of principal Gurva« ture at the point (r cos 9, r sin B, t) are given by the equation *'($)'*-»S!''*(»)T-!-(S)T- I. Solution by the Rev. J. L. Eitohin, M.A. The radii of principal curvature are given by the equation + (l+/i«+3^-0 (1). The general equation of a conoid, axis of surface axis of ir, is z » ^ f^ ^ and since w^rco&9, y « r sin 9, this becomes z^^ (tan 9) »/(9) ; there- fore z is independent of r. Hence we find — n -. — Sin 9 — , — as -- cos 9 — ; dx r aO ay r dJ9 andhence !+;.+,« .1+ J (g)^. J {^+ g)'}; also ^*,-^Asirie^^-.^|^lsinel'^*:-lfisine^^^ da^ dx\r d9) dr \r d») dx d9 \r d») dx Effecting the differentiations and reducmg, remembering tha€ — is a func- d9 tion of 9 only, we get Digitized by VjOOQIC 112 ^z 8in»tf <P2 2 sin 2B dz ^ ^ , . ., , 6?z coR^ $ cPz 2sm2e dz . ..m.lariy Tj^' -^ ^^ l^H'*' and ^ = _C2£^^ + !!!i|?^V hence '.'-'-M=tS*""S?.-"-»(S)'I 4 fdz\^ "";iU; We find also that (1 +p2) «-2py*+ (1 + g») ri -. i ti. Hence (1) becomes _ 4 /d«y ,_ 1 d^ J \d9) i \ \dej ) ?\de)^ f^^d^ r r* '^(S)-si-*(S)T-i-'(i)'r- = 0, II. Solution by the Pbofoses. Let the increments of r» 9, z, at a point Q near the ^ven point P, be A89, 80, Zz ; then the tangent plane at P is the limiting position of the plane containing the generating line Xsin 9 s= Ycos 9, Z^z, and passing through an adjacent point of the surface; its equation being taken Xsintf—YcosO + Je (Z-«) = 0, ir is the Umit of (^ + ^)""^^^ or r -^ ^. But if p be the 8z dO radius of curvature of the normal section containing the ultimate position of the line PQ, and p be the perpendicular from Q on the tangent plane at P, 2p « limit of ^, V Now PQ2 « x'aea + ^28^ + 5^2, -(r + XW)^sinW+rf^M+f?f???+....^ ^ ^ de \de de- z J ^^ p r. — r.,z\h^ Co /dz\^ih V whence '^ " V "^ UJ » "^ dzy de) + X2 ^ d0^ de and the principal radii of curvature are the maximum and minimum values of this for difi'erent vhIucs of A., and are therefore given by the-equation Printed by C. F. Hodgson & Son, Gough Square, Fleet Street. Digitized byLnOOQlC Digitized byLnOOQlC Digitized byVjOOQlC Digitized byVjOOQlC Digitized by Google